RD Sharma XI 2019 Solutions for Class 11 Science Math Chapter 19 Arithmetic Progressions are provided here with simple step-by-step explanations. These solutions for Arithmetic Progressions are extremely popular among class 11 Science students for Math Arithmetic Progressions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XI 2019 Book of class 11 Science Math Chapter 19 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XI 2019 Solutions. All RD Sharma XI 2019 Solutions for class 11 Science Math are prepared by experts and are 100% accurate.

Page No 19.11:

Question 1:

Find:
(i) 10th term of the A.P. 1, 4, 7, 10, ...
(ii) 18th term of the A.P. 2, 32, 52, ...
(iii) nth term of the A.P. 13, 8, 3, −2, ...

Answer:

(i) 1, 4, 7, 10...

We have:
a = 1
d = 4-1 = 3

a10 =a+(10-1)d                  an=a+n-1d       = a +9d        = 1+9×3        = 28

(ii) 2, 32, 52...
We have:
a = 2d = 32 - 2 = 22

a18  = a+18-1d     an=a+n-1d         = a+17d        = 2+1722        = 2+342        = 35 2

(iii) 13, 8, 3, −2...
We have:
a = 13d = 8 -13 = -5

an = a+(n-1)d     = 13+(n-1)-5     = 13-5n+5     = 18-5n



Page No 19.12:

Question 2:

If the sequence < an > is an A.P., show that
am +n +amn = 2am.

Answer:

Let the sequence < an >  be an A.P. with the first term being A and the common difference being D.
To prove: am +n +amn = 2am

LHS: am +n +amn

= A+(m+n-1)D + A+(m-n-1)D     {an = a+(n-1)d}= A+mD+nD-D +A+mD-nD-D=2A+2mD-2D   ...(i)


RHS: 2am

= 2[A+(m-1)D]= 2A +2mD-2D   ...(ii)

From (i) and (ii), we get:
LHS = RHS
Hence, proved.

Page No 19.12:

Question 3:

(i) Which term of the A.P. 3, 8, 13, ... is 248?
(ii) Which term of the A.P. 84, 80, 76, ... is 0?
(iii) Which term of the A.P. 4, 9, 14, ... is 254?

Answer:

(i) 3, 8, 13...
Here, we have:
a = 3
d = 8-3  =5
Let an = 248 a+n-1d = 2483+n-15 = 248n-15 =245n-1 = 49 n= 50

Hence, 248 is the 50th term of the given A.P.

(ii) 84, 80, 76...
Here, we have:
a = 84
d = 80-84 = -4
Let an =0a+(n-1)d = 084 + (n-1)-4 = 0 (n-1)-4 = -84(n-1) = 21 n = 22

Hence, 0 is the 22nd term of the given A.P.

(iii) 4, 9, 14...
Here, we have:
a = 4
d = 9-4 = 5
Let an = 254a+n-1 d = 2544+n-1 5 = 254n-1 5 =250n-1 = 50 n= 51

Hence, 254 is the 51st term of the given A.P.

Page No 19.12:

Question 4:

(i) Is 68 a term of the A.P. 7, 10, 13, ...?
(ii) Is 302 a term of the A.P. 3, 8, 13, ...?

Answer:

(i) 7, 10, 13...
Here, we have:
a = 7
d = 10-7 = 3Let an = 68 a+(n-1) d = 687+(n-1)(3) = 68(n-1)(3) = 61(n-1) = 613n =  613 +1 = 643

Since n is not a natural number.So, 68 is not a term of the given A.P.

(ii) 3, 8, 13...
Here, we have:
a  = 3
d = 8-3=5Let an = 302 a+n-1d = 3023+n-15 = 302n-15 =299n-1 = 2995n = 2995+1 = 3045
Since n is not a natural number.So, 302 is not a term of the given A.P.

Page No 19.12:

Question 5:

(i) Which term of the sequence 24, 2314, 2212, 2134 ... is the first negative term?
(ii) Which term of the sequence 12 + 8i, 11 + 6i, 10 + 4i, ... is (a) purely real (b) purely imaginary?

Answer:

(i)  24, 2314, 2212, 2134...
This is an A.P.
Here, we have:
a = 24
d = 2314- 24 = -34Let the first negative term be an.Then, we have: an <0a+n-1 d<024+n-1  -34<024-3n4+34<024+34<3n4994<3n499<3nn >33

Thus, the 34th term is the first negative term of the given A.P.

(ii)
(a) 12 + 8i, 11 + 6i, 10 + 4i...
This is an A.P.
Here, we have:
a = 12 + 8i
d = 11+6i - 12-8i   = -1-2iLet the real term be an = a+n-1d. an = 12+8i+n-1-1-2i       =  12+8i+ -n+1-2in+2i       = 12+8i-n+1-2in+2i       = 13-n +8-2n+2i       =  13-n +10-2nian has to be real.10-2n = 0 n= 5  

(b) 12 + 8i, 11 + 6i, 10 + 4i...
This is an A.P.
Here, we have:
a = 12 + 8i
d = 11+6i - 12-8i   = -1-2iLet the imaginary term be an = a+n-1d an = 12+8i+n-1-1-2i       =  12+8i+ -n+1-2in+2i       = 12+8i-n+1-2in+2i       = 13-n +8-2n+2i       =  13-n +10-2nian has to be imaginary. 13-n = 0 n=13 

Page No 19.12:

Question 6:

(i) How many terms are there in the A.P.
7, 10, 13, ... 43?
(ii) How many terms are there in the A.P.
-1,-56,-23,-12, ...,103?

Answer:

(i) 7, 10, 13...43
Here, we have:
= 7
d = (10-7) = 3an = 43
Let there be n terms in the given A.P.

Also, an  =a+n-1d43 = 7+n-1336 = n-1312 = n-113 = n

Thus, there are 13 terms in the given A.P.

(ii) -1,-56,-23,-12, ...,103

Here, we have:
= -1
d = -56--1 = 1-56= 16an = 103
Let there be n terms in the given A.P.

Also, an  =a+n-1d103= -1+n-116133 = n-11626= n-127 = n

Thus, there are 27 terms in the given A.P.

Page No 19.12:

Question 7:

The first term of an A.P. is 5, the common difference is 3 and the last term is 80; find the number of terms.

Answer:

 Here, a = 5, d = 3, an = 80
Let the number of terms be n.
Then, we have:

an = a+n-1d80 = 5+n-1375 = n-1325 =  n-126 = n

Thus, there are 26 terms in the given A.P.

Page No 19.12:

Question 8:

The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.

Answer:

Given:

a6 = 19 a+6-1d = 19               an=a+n-1d a+5d=19           ..1And,  a17 = 41a+17-1d=41               an=a+n-1da+16d = 41          ..2Solving the two equations, we get,16d-5d = 41-1911d = 22d = 2   Putting d = 2 in the eqn 1, we get:a+5×2 = 19a = 19-10a= 9   


We know:
a40  =a+40-1d         an=a+n-1d       = a+39d       = 9+39×2              = 9+78 = 87

Page No 19.12:

Question 9:

If 9th term of an A.P. is zero, prove that its 29th term is double the 19th term.

Answer:

Given:
a9 = 0           a+9-1d=0       an=a+n-1da+8d = 0a = -8d   ...(i)

To prove:
a29 = 2a19

Proof:
LHS: a29 =a+29-1d= a+28d=-8d+28d    From(i)= 20d

RHS: 2a19=2a+19-1d= 2(a+18d)= 2a+36d= 2(-8d)+36d       From(i)= -16d+36d= 20d

LHS = RHS
Hence, proved.

Page No 19.12:

Question 10:

If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of the A.P. is zero.

Answer:

Given:
10a10 = 15a1510a+10-1d=15a+15-1d10(a+9d) = 15(a+14d)10a+90d = 15a+210d0 = 5a+120d0 = a+24da = -24d   ...(i)

To show:
a25 = 0LHS: a25=a+25-1d =a+24d=-24d +24d         From(i)= 0  = RHSHence, proved.

Page No 19.12:

Question 11:

The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.

Answer:

Given:

 a10 = 41a+10-1d=41       an=a+n-1da+9d = 41 And, a18=73a+18-1d=73         an=a+n-1da+17d =73 Solving the two equations, we get:17d-9d = 73-418d = 32d = 4               ...(i)Putting the value in first equation, we get:a+9×4 = 41a+36 = 41a = 5                 ...(ii)

 a26 = a+26-1d      an=a+n-1d a26 = a+25d a26 =5+25×4          From (i) and (ii)a26 =5+100 = 105

Page No 19.12:

Question 12:

In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.

Answer:

Given:
a24 = 2a10a+24-1d=2a+10-1da+23d = 2(a+9d)a+23d = 2a+18d5d = a       ...(i)

To prove:a72 = 2a34LHS: a72 =a+72-1da72 =a+71da72 =5d+71d     From(i)a72 =76d  

RHS: 2a34 = 2a+34-1d2a34 =2 a+33d2a34 =2(5d+33d)            Form(i)2a34 = 238d2a34 =76d
∴ RHS = LHS
Hence, proved.

Page No 19.12:

Question 13:

If (m + 1)th term of an A.P. is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n + 1)th term.

Answer:

Given:
am+1 = 2an+1a+(m+1-1)d = 2[a+(n+1-1)d]a+md = 2(a+nd)a+md = 2a+2nd0 = a+2nd-md    nd=md-a2          ...(i)

To prove:

a3m+1 = 2am+n+1LHS: a3m+1 =a+(3m+1-1)da3m+1 =a+3mdRHS: 2am+n+1 = 2[a+(m+n+1-1)d]2am+n+1 =2(a+md+nd)2am+n+1 = 2a+md+md-a2     From(i)2am+n+1 = 22a+2md+md-a2

2am+n+1=2a+3md22am+n+1=a+3md

∴ LHS = RHS

Page No 19.12:

Question 14:

If the nth term of the A.P. 9, 7, 5, ... is same as the nth term of the A.P. 15, 12, 9, ... find n.

Answer:

Given:
nth term of the A.P. 9, 7, 5... is the same as the nth term of the A.P. 15, 12, 9...

Considering 9, 7, 5...a = 9, d= 7-9 = -2nth term = 9+(n-1)(-2)           an=a+n-1d               = 9-2n+2               = 11-2n         ...(i)

Considering 15, 12, 9, ...a = 15, d = 12-15 = -3nth term  =15+(n-1)(-3)   an=a+n-1d               = 15-3n+3               = 18-3n ...(ii)

Equating (i) and (ii), we get:
11-2n = 18-3nn = 7

Thus, 7th terms of both the A.P.s are the same.

Page No 19.12:

Question 15:

Find the 12th term from the end of the following arithmetic progressions:
(i) 3, 5, 7, 9, ... 201
(ii) 3, 8, 13, ..., 253
(iii) 1, 4, 7, 10, ..., 88

Answer:

(i) 3, 5, 7, 9...201
Consider the given progression with 201 as the first term  and −2 as the common difference.
12th term from the end = 201+(12-1)(-2) = 179

(ii) 3, 8, 13...253
Consider the given progression with 253 as the first term  and −5 as the common difference.
12th term from the end = 253+(12-1)(-5) = 198

(iii) 1, 4, 7, 10...88
Consider the given progression with 88 as the first term and −3 as the common difference.
12th term from the end = 88+(12-1)(-3) = 55

Page No 19.12:

Question 16:

The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference.

Answer:

Given:
Let the first term of the A.P. be a and the common difference be d.
a4=3a a+4-1d = 3a a+3d = 3a3d = 2aa=3d2     ...(i)And, a7-2a3=1a+7-1d-2a+3-1d=1a+6d -2(a+2d)  =1a+6d-2a-4d =1-a+2d = 1-3d2+2d = 1        From (i)   -3d+4d2=1d2=1d=2
Putting the value in (i), we get:a=3×22a = 3

Page No 19.12:

Question 17:

Find the second term and nth term of an A.P. whose 6th term is 12 and the 8th term is 22.

Answer:

Given:
a6 =12a+6-1d=12a+5d = 12              ...(i)a8  =22a+8-1d=22a+7d = 22              ...(ii)Solving (i) and (ii), we get:2d = 10d = 5Putting the value of d in (i), we get: a+5×5 = 12a=12-25 = -13 a2 = a+2-1d=a+d = -13+5 = -8Also, an = a+(n-1)d                 =-13+(n-1)5               = -13+5 n-5                = 5n-18 

Page No 19.12:

Question 18:

How many numbers of two digit are divisible by 3?

Answer:

The two digit numbers that are divisible by 3 are:
12, 15, 18...96, 99
This is an A.P. whose first term is 12 and the common difference is 3.
We have:an = 9912+(n-1)3 = 99(n-1)3 =87(n-1) = 29n = 30
Thus, there are 30 such terms.

Page No 19.12:

Question 19:

An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32nd term.

Answer:

Given:
a = 7, n= 60, l= 125
l = a+(n-1)d125  =7+(60-1)d125 = 7+59d118 = 59d2 = d

a32 = a+32-1d       =a+31d        = 7+31×2        = 7+62        = 69

Page No 19.12:

Question 20:

The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of the A.P.

Answer:

Let a be the first term and d be the common difference. Then,
a4+a8 = 24a+4-1d+a+8-1d=24a+3d+a+7d = 242a+10d = 24  a+5d = 12      ...(i)Also, a6+a10   =34a+6-1d+a+10-1d=34a+5d+a+9d = 342a+14d = 34a+7d = 17    ...(ii)Solving (i) and (ii), we get:2d = 5d = 52Substituing the value in (i), we get:a+552 = 12a+252 = 12a = -12

Page No 19.12:

Question 21:

How many numbers are there between 1 and 1000 which when divided by 7 leave remainder 4?

Answer:

A number N is divided by 7 leaves a remainder 4.

N = 7q + 4

N can take values 4, 11, 18, ..... 998

Now,
4, 11, 18, ..... 998 are in arithmetic progression.
First term a = 4
common difference d = 7
last term l = 998

We know that,
l = a + (n − 1)d
⇒ 998 = 4 + (n − 1)7
⇒ 998 = 4 + 7n − 7
⇒ 998 = 7n − 3
⇒ 1001 = 7n
n=10017
n = 143

Hence, 143 numbers are there between 1 and 1000 which when divided by 7 leave remainder 4.

Page No 19.12:

Question 22:

The first and the last terms of an A.P. are a and l respectively. Show that the sum of nth term from the beginning and nth term from the end is a + l.

Answer:

Given:
First term = a
Last term = l
nth term from the beginning  = a+(n-1)d, where d is the common difference.
nth term from the end = l+(n-1)(-d) =l-dn+d
Their sum = a+(n-1)d +l-dn+d
= a+nd-d+l-nd+d = a+l
Hence, proved.

Page No 19.12:

Question 23:

If < an > is an A.P. such that a4a7=23, finda6a8.

Answer:

Given:
< an > is an A.P.
a4a7 = 23a+4-1da+7-1d= 23     a+3da+6d= 233(a+3d) = 2(a+6d) 3a+9d = 2a+12da = 3d ....(i)

 a6a8 = a+6-1da+8-1da6a8 =a+5da+7da6a8 =3d+5d3d+7d       From(i)a6a8 = 8d10da6a8 = 4d5d=45

Page No 19.12:

Question 24:

If θ1, θ2, θ3, ..., θn are in AP, whose common difference is d, then show thatsecθ1secθ2+secθ2secθ3+...+secθn-1secθn=tanθn-tanθ1sind                                                                                                   NCERTEXEMPLAR

Answer:

As, θ1, θ2, θ3, ..., θn are in APSo, d=θ2-θ1=θ3-θ2=...=θn-θn-1       .....iNow,LHS=secθ1secθ2+secθ2secθ3+...+secθn-1secθn=1cosθ1cosθ2+1cosθ2cosθ3+...+1cosθn-1cosθn=1sindsindcosθ1cosθ2+sindcosθ2cosθ3+...+sindcosθn-1cosθn=1sindsinθ2-θ1cosθ1cosθ2+sinθ3-θ2cosθ2cosθ3+...+sinθn-θn-1cosθn-1cosθn                             Using i=1sindsinθ2cosθ1-cosθ2sinθ1cosθ1cosθ2+sinθ3cosθ2-cosθ3sinθ2cosθ2cosθ3+...+sinθncosθn-1-cosθnsinθn-1cosθn-1cosθn=1sindsinθ2cosθ1cosθ1cosθ2-cosθ2sinθ1cosθ1cosθ2+sinθ3cosθ2cosθ2cosθ3-cosθ3sinθ2cosθ2cosθ3+...+sinθncosθn-1cosθn-1cosθn-cosθnsinθn-1cosθn-1cosθn=1sindtanθ2-tanθ1+tanθ3-tanθ2+...+tanθn-tanθn-1=1sind-tanθ1+tanθn=tanθn-tanθ1sind=RHS



Page No 19.15:

Question 1:

The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.

Answer:

Let the three terms of the A.P. be a-d, a, a+d.Then, we have:a-d+a+a+d = 213a =21a = 7   ....(i)Also, (a-d)(a+d) - a = 6a2-d2 -a = 649-d2 -7 = 636=d2±6 = dWhen d=6, a=7, we get:1, 7, 13 When d=-6, a=7, we get:13, 7, 1

Page No 19.15:

Question 2:

Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers.

Answer:

Let the three numbers be a-d, a, a+d.Their sum = 27 a-d+a+a+d = 273a = 27a=9   ...(i)Product = (a-d)a(a+d)  =648 a(a2-d2) = 6489(81-d2)  =648(81-d2)  =72d2 = 9d = ±3When a=9, d=3, we have:6, 9, 12When a=9, d=-3, we have:12, 9, 6

Page No 19.15:

Question 3:

Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.

Answer:

Let the four numbers be a-3d, a-d, a+d, a+3d.Sum = 50 a-3d+ a-d+a+d+a+3d = 504a =50a = 25 2 ...(i)Also, a+3d = 4(a-3d)a+3d  =4a-12d3a = 15da = 5d252×5 = d    Using (i)52=dSo, the terms are as follows: 25 2-3×52, 25 2-52,25 2+52, 25 2+3×52= 5, 10, 15, 20

Page No 19.15:

Question 4:

The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.

Answer:

Let the numbers be (a-d), a, (a+d).Sum = a-d+a+a+d =123a = 12a  =4Also, (a-d)3+a3+(a+d)3 =288a3-d3-3a2d+3ad2+a3+a3+d3+3a2d+3ad2 =2883a3+6ad2 = 288343+6×4×d2 = 288192+24d2  =28824d2  =96d2 = 4d = ±2When a=4,  d=2, the numbers are 2, 4, 6.When a=4,  d=-2, the numbers are 6, 4, 2.

Page No 19.15:

Question 5:

If the sum of three numbers in A.P. is 24 and their product is 440, find the numbers.

Answer:

Let the three numbers be (a-d), a, (a+d).Sum = 24(a-d)+a+(a+d)  =243a  =24a = 8   ...(i)Product =a(a-d)(a+d) =440a(a2-d2) = 4408(64-d2)  =440        Form (i)(64-d2)  =55d2 = 9d = ±3

With a = 8, d = 3,  we have:5, 8, 11With a = 8, d =-3,  we have:11, 8, 5

Page No 19.15:

Question 6:

The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the angles.

Answer:

Let the angles be (A)°, (A+d)°, (A+2d)°, (A+3d)°.
Here, = 10
So, A°, (A+10)°, (A+20)°, (A+30)° are the angles of a quadrilateral whose sum is 360o.

 A°, (A+10)°, (A+20)°, (A+30)°= 360°4A = 360-60A = 3004=75°

The angles are as follows:75°, (75+10)°, (75+20)°, (75+30)°, i.e. 75°, 85°,95°, 105°



Page No 19.30:

Question 1:

Find the sum of the following arithmetic progressions:
(i) 50, 46, 42, ... to 10 terms
(ii) 1, 3, 5, 7, ... to 12 terms
(iii) 3, 9/2, 6, 15/2, ... to 25 terms
(iv) 41, 36, 31, ... to 12 terms
(v) a + b, ab, a − 3b, ... to 22 terms
(vi) (xy)2, (x2 + y2), (x + y)2, ... to n terms
(vii) x-yx+y,3x-2yx+y,5x-3yx+y, ... to n terms

Answer:

(i) 50, 46, 42 ... to 10 terms
We have: a = 50, d = 46-50 =-4n = 10Sn = n22a+(n-1)d= 1022×50+(10-1)(-4)= 5100-36 = 320

(ii) 1, 3, 5, 7 ... to 12 terms
We have: a = 1, d = 3-1 =2n = 12Sn = n22a+(n-1)d= 1222×1+(12-1)(2)= 624 = 144

(iii) 3, 9/2, 6, 15/2 ... to 25 terms
We have: a = 3,  d = 9/2-3 =3/2n = 25Sn = n22a+(n-1)d= 2522×3+(25-1)(3/2)=252×42= 525

(iv) 41, 36, 31 ... to 12 terms
We have: a = 41, d = 36-41 =-5n = 12Sn = n22a+(n-1)d= 1222×41+(12-1)(-5)= 6×27= 162

(v) a + b, ab, a − 3b ... to 22 terms
We have:First term = a+b, d = a-b-a-b =-2bn =22Sn = n22a+(n-1)d= 2222×(a+b)+(22-1)(-2b)= 112a-40b = 22a-440b

(vi) (xy)2, (x2 + y2), (x + y)2 ... to n terms
We have: a = (x − y)2, d = x2+y2-(x − y)2  =2xySn = n22a+(n-1)d= n22(x − y)2+(n-1)(2xy)= n2×2(x − y)2+(n-1)(xy)= n(x − y)2+(n-1)(xy)

(vii) x-yx+y,3x-2yx+y,5x-3yx+y ... to n terms
We have: a = x-yx+y, d =3x-2yx+y-x-yx+y=2x-yx+ySn = n22a+(n-1)d= n22x-yx+y+(n-1)2x-yx+y=n2(x+y)(2x-2y)+(2x-y)(n-1)=n2(x+y)2x-2y-2x+y+n(2x-y)=n2(x+y)n(2x-y)-y

Page No 19.30:

Question 2:

Find the sum of the following series:
(i) 2 + 5 + 8 + ... + 182
(ii) 101 + 99 + 97 + ... + 47
(iii) (ab)2 + (a2 + b2) + (a + b)2 + ... + [(a + b)2 + 6ab]

Answer:

(i) 2 + 5 + 8 + ... + 182
Here, the series is an A.P. where we have the following:
a = 2d=5-2 = 3an=1822+(n-1)(3)=1822+3n-3=1823n-1=1823n=183n=61 Sn =n22a+(n-1)dS61= 6122×2+61-1×3            =6122×2+60×3           = 5612

(ii) 101 + 99 + 97 + ... + 47
Here, the series is an A.P. where we have the following:
a = 101d=99-101 = -2an=47101+(n-1)(-2)=47101-2n+2=472n-2=542n=56n=28Sn =n22a+(n-1)d S28 = 2822×101+28-1×(-2)             = 2822×101+27×(-2)               = 2072

(iii) (ab)2 + (a2 + b2) + (a + b)2 + ... + [(a + b)2 + 6ab]
Here, the series is an A.P. where we have the following:
a = (a-b)2d=a2+b2- (a-b)2 = 2aban=[(a + b)2+6ab](a-b)2+(n-1)(2ab)= [(a + b)2+6ab] a2+b2-2ab+2abn-2ab=[a2+b2+2ab+6ab] a2+b2-4ab+2abn=a2+b2+8ab 2abn=12ab n=6 Sn  =n22a+(n-1)dS6= 622(a-b)2+6-1 2ab           = 32(a2+b2-2ab)+10ab          = 32a2+2b2-4ab+10ab          = 32a2+2b2+6ab          = 6a2+b2+3ab

Page No 19.30:

Question 3:

Find the sum of first n natural numbers.

Answer:

The first n natural numbers are:
1, 2, 3, 4...
a = 1, d = 1, Total terms = n

Sn = n22a+(n-1)dSn = n22×1+(n-1)1Sn = n22+(n-1)1Sn = n2n+1

Page No 19.30:

Question 4:

Find the sum of all natural numbers between 1 and 100, which are divisible by 2 or 5.

Answer:

We have to find the sum of all the natural numbers that are divisible by 2 or 5
Required Sum = Sum of the natural numbers between 1 and 100 that are divisible by 2 + Sum of the natural numbers between 1 and 100 that are divisible by 5
                      − Sum of the natural numbers between 1 and 100 that are divisible by 2 and 5, i.e by 10
             
                     =2+4+6+8+...+98+5+10+15+...+95-10+20+30+...+90=5022+98+2025+95-10210+90=2500+1000-500=3000

Page No 19.30:

Question 5:

Find the sum of first n odd natural numbers.

Answer:

The first n odd natural numbers are:
1, 3, 5, 7, 9...
a = 1, d = 2, Total terms = n

Sn = n22a+(n-1)dSn = n22×1+(n-1)2Sn = n22+(n-1)2Sn = n22nSn = n2

Page No 19.30:

Question 6:

Find the sum of all odd numbers between 100 and 200.

Answer:

All the odd numbers between 100 and 200 are:
101, 103...199
Here, we have:
a = 101d = 2an = 199101+(n-1)×2=1992n-2=982n = 100n = 50 Sn  = n22a+(n-1)d S50 = 5022×101+(50-1)2S50 = 25202+98
S50=7500

Page No 19.30:

Question 7:

Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.

Answer:

The odd integers between 1 and 1000 that are divisible by 3 are:
3, 9, 15, 21...999
Here, we have:
a = 3, d = 6an= 9993+(n-1)6=9993+6n-6=9996n=1002n =167 Sn = n22a+(n-1)dS167 = 16722×3+(167-1)6S167 = 16721002 = 83667Hence, proved.

Page No 19.30:

Question 8:

Find the sum of all integers between 84 and 719, which are multiples of 5.

Answer:

The integers between 84 and 719, which are multiples of 5 are:
85, 90...715
Here, we have:
a = 85d= 5an= 71585+(n-1)5=715 5n-5=630n=127 Sn = n22a+(n-1)d S127 = 12722×85+(127-1)5S127 = 1272800 = 50800



Page No 19.31:

Question 9:

Find the sum of all integers between 50 and 500 which are divisible by 7.

Answer:

The integers between 50 and 500 that are divisible by 7 are:
56, 63...497
Here, we have:
a = 56d =7 an = 49756+(n-1)7=4977n-7=4417n=448n=64 Sn = n22a+(n-1)d S64 = 6422×56+(64-1)7 S64 = 322×56+63×7 = 17696

Page No 19.31:

Question 10:

Find the sum of all even integers between 101 and 999.

Answer:

The even integers between 101 and 999 are:
102, 104...998
Here, we have:
a= 102d = 2 an = 998102+(n-1)2=9982n-2=8962n = 898n=449Sn = n22a+(n-1)dS449 = 44922×102+(449-1)×2S449 = 44921100S449 = 246950

Page No 19.31:

Question 11:

Find the sum of all integers between 100 and 550, which are divisible by 9.

Answer:

The integers between 100 and 550 that are divisible by 9 are:
108, 117...549
Here, we have:
a= 108 d = 9an = 549108+(n-1)(9)=5499n-9=4419n=450n=50Sn = n22a+(n-1)dS50 = 5022×108+(50-1)×9S50 = 25657 = 16425

Page No 19.31:

Question 12:

Find the sum of the series:
3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 + ... to 3n terms.

Answer:

The given sequence i.e., 3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 +..... to 3n terms.

can be rewritten as 3 + 6 + 9 + .... to n terms + 5 + 9 + 13 + .... to n terms + 7 + 12 + 17 + .... to n terms

Clearly, all these sequence forms an A.P. having n terms with first terms 3, 5, 7 and common difference 3, 4, 5

Hence, required sum = n22×3+n-13+n22×5+n-14+n22×7+n-15

                                  =n26+3n-3+10+4n-4+14+5n-5=n212n+18=3n2n+3

Page No 19.31:

Question 13:

Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7.

Answer:

The sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7 are:
103, 119...791
Here, we have:
a = 103
d = 16
an = 791
We know:an = a+(n-1)d791 = 103+(n-1)×16688 = 16n-16704 = 16n44 = nAlso, Sn = n2[2a+(n-1)d]S44 = 442[2×103+(44-1)×16]S44 = 22 [206+688]S44 = 22 × 894 = 19668

Page No 19.31:

Question 14:

Solve: (i) 25 + 22 + 19 + 16 + ... + x = 115
(ii) 1 + 4 + 7 + 10 + ... + x = 590.

Answer:

(i) 25 + 22 + 19 + 16 + ... + x = 115
Here, a = 25, d = -3, Sn = 115
We know:Sn = n22a+(n-1)d115 = n22×25+(n-1)×-3115×2 = n50-3n+3230 = n53-3n230 = 53n - 3n23n2-53n+230 = 0By quadratic formula: n = -b±b2-4ac2aSubstituting a=3, b=-53 and c=230, we get:n= 53±532-4×3×2302×3 = 466, 10n = 10, as n466 an= a+(n-1)d x = 25+(10-1)(-3)x =25-27 = -2

(ii) 1 + 4 + 7 + 10 + ... + x = 590
Here, a = 1, d = 3,
We know:Sn = n22a+(n-1)d590 = n22×1+(n-1)×3590×2 = n2+3n-31180 = n3n-11180 = 3n2 - n3n2-n-1180 = 0By quadratic formula: n = -b±b2-4ac2aSubstituting a=3, b=-1 and c=-1180, we get:n= 1±12+4×3×11802×3 = -1186, 20n = 20, as n-1186 an=  x = a+(n-1)d x = 1+(20-1)(3)x =1+60-3 = 58

Page No 19.31:

Question 15:

Find the rth term of an A.P., the sum of whose first n terms is 3n2 + 2n.                                                                    [NCERT EXEMPLAR]

Answer:

Let a and d be the first term and the common difference of the given A.P., respectivelyAs, Sn=3n2+2nSo, a=S1=3×12+2×1=3+2=5 andS2=3×22+2×2=12+4=16a+a2=16a+a+d=162a+d=162×5+d=16d=16-10d=6Now,ar=a+r-1d=5+r-1×6=5+6r-6 ar=6r-1

Page No 19.31:

Question 16:

How many terms are there in the A.P. whose first and fifth terms are −14 and 2 respectively and the sum of the terms is 40?

Answer:

We have: a = -14 and Sn = 40   ...(i)a5 = 2a+5-1d = 2-14+4d = 24d = 16d = 4                          ...(ii)Also, Sn = n22a+(n-1)d40 = n22-14+(n-1)×4     (From(i) and (ii))80 = n-28+4n-480 = 4n2-32nn2-8n-20 = 0(n-10)(n+2) = 0 n = 10, -2But, n cannot be negative. n = 10 

Page No 19.31:

Question 17:

The sum of first 7 terms of an A.P. is 10 and that of next 7 terms is 17. Find the progression.

Answer:

We have: S7 = 10722a+(7-1)d = 10722a+6d  =10a+3d = 107   ...(i)Also, the sum of the next seven terms = S14 - S7 = 171422a+14-1d-722a+(7-1)d = 1772a+13d-722a+6d  =1714a+91d - 7a-21d = 177a+70d = 17a+10d = 177   ...(ii)From (i) and (ii), we get:107 -3d = 177-10d7d= 1d = 17Putting the value in (i), we get:a+3d = 107a+37 = 107a = 1 a = 1, d = 17
The progression thus formed is 1, 87, 97, 107...

Page No 19.31:

Question 18:

The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.

Answer:

Given: a3 = 7, a7 - 3a3 = 2We have: a3 = 7a+3-1d=7 a+2d  =7             ...(i) Also, a7 - 3a3 = 2a7 -21 = 2  (Given)a+7-1d=23a+6d = 23            ...(ii)From (i) and (ii), we get:4d = 16d = 4Putting the value in (i), we get: a+2(4) = 7a = -1S20 = 2022-1+20-1(4)S20 = 10-2+76S20 = 1074  =740 a  =-1, d = 4, S20 = 740

Page No 19.31:

Question 19:

The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.

Answer:

Given: a = 2, l = 50, Sn =442Now, Sn =442n2a+l = 442n22+50 = 442n = 17  a17=50a+(17-1)×d  = 50  2+16d = 50d = 3

Page No 19.31:

Question 20:

The number of terms of an A.P. is even; the sum of odd terms is 24, of the even terms is 30, and the last term exceeds the first by 10 12, find the number of terms and the series.

Answer:

Let total number of terms be 2n.
According to question, we have:

a1+a3+...+a2n-1  =24  ...(1)a2+a4+...+a2n  = 30    ...(2)Subtracting (1) from (2), we get:d+d+...+upto n terms = 6nd = 6                               ...(3)Given: a2n= a1+212 a2n- a1 = 212a+(2n-1)d -a = 212    [ a2n = a+(2n-1)d, a1 =a]2nd-d = 2122×6 - d = 212      From(3)d = 32Putting the value in (3), we get:n = 42n = 8Thus, there are 8 terms in the progression. 

To find the value of the first term:a2+a4+...+a2n = 30(a+d) +(a+3d)+...+[a+(2n-1)d] = 30n2a+d+a+(2n-1)d = 30Putting n=4 and d=32, we get: a = 32So, the series will be 112, 3, 412...

Page No 19.31:

Question 21:

If Sn = n2p and Sm = m2p, mn, in an A.P., prove that Sp = p3.

Answer:

Sn = n2pn22a+(n-1)d = n2p2np = 2a+(n-1)d     ...(i)Sm = m2pm22a+(m-1)d =  m2p2mp = 2a+(m-1)d    ...(ii)Subtracting (ii) from (i), we get:2p(n-m) = (n-m)d2p = d                         ...(iii)Substituing the value in (i), we get:nd = 2a+(n-1)dnd -nd+d = 2aa = d2= p from(iii)      ...(iv) Sp = p22a+p-1dSp = p22p+p-12pSp = p22p+2p2-2pSp = p22p2Sp = p3

Page No 19.31:

Question 22:

If 12th term of an A.P. is −13 and the sum of the first four terms is 24, what is the sum of first 10 terms?

Answer:

Let a be the first term and d be the common difference.
 a12 = -13a+12-1d =-13a+11d =-13      ...(i)Also, S4 = 24422a+(4-1)d = 2422a+3d = 242a+3d = 12       ...(ii) From (i) and (ii), we get:19d=-38d = -2Putting the value of d in (i), we get:a+11-2=-13a=9S10 = 1022×9+(10-1)-2 S10 = 518+9-2 = 0

Page No 19.31:

Question 23:

If the 5th and 12th terms of an A.P. are 30 and 65 respectively, what is the sum of first 20 terms?

Answer:

We have:a5 = 30a+5-1d = 30a+4d = 30   ...(i)Also, a12 = 65a+12-1d = 65a+11d = 65     .....(ii)Solving (i) and (ii), we get:7d=35d=5Putting the value of d in (i), we get:a+4×5=30a=10 S20  = 2022×10+(20-1)×5S20  = 102×10+(20-1)×5S20  =1150

Page No 19.31:

Question 24:

Find the sum of n terms of the A.P. whose kth terms is 5k + 1.

Answer:

Given: ak = 5k+1For k=1, a1 = 5×1+1 = 6For k =2, a2 =5×2+1 = 11For k = n, an = 5n+1 Sn = n2a+anSn = n26+5n+1 = n25n+7

Page No 19.31:

Question 25:

Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

Answer:

The two-digit numbers which when divided by 4 yield 1 as remainder are 13, 17....97.

 a =13, d = 4, an = 97 an = a+(n-1)d97 = 13+(n-1)484 = 4n-488 = 4n22= n   ...1Also, Sn = n22a+(n-1)d S22 = 2222×13+(22-1)×4        (From1)S22 = 11110 = 1210

Page No 19.31:

Question 26:

If the sum of a certain number of terms of the AP 25, 22, 19, ... is 116. Find the last term.

Answer:

The given A.P. is 25,.22,.19.....
Here, a = 25, d = 22-25=-3
Sn  =116n22a+n-1d = 116n2×25+n-1-3 = 23250n-3n2+3n = 2323n2 -53n+232 = 0

3n2-29n-24n+232 = 0n(3n-29)-8(3n-29)  =0(3n-29)(n-8) = 0n = 293 or 8Since n cannot be a fraction, n = 8.Thus, the last term:an = a+(n-1)da8 =25+8-1×-3a8 = 4

Page No 19.31:

Question 27:

Find the sum of odd integers from 1 to 2001.

Answer:

The odd integers from 1 to 2001 are 1, 3, 5.....2001.It is an AP with a=1 and d=2.an = 20011+(n-1)2=20012n-2=20002n=2002n=1001Also, S1001 = 100122×1+1001-12S1001 = 100122×1+10002S1001 =10012×2002 = 1002001

Page No 19.31:

Question 28:

How many terms of the A.P. −6, -112, −5, ... are needed to give the sum −25?

Answer:

Given:AnA.P. with a =-6 and d = -112--6 = 12Sn = -25 -25 = n22×-6+n-112-25 = n2-12+n2-12-50 = nn2-252-100 = nn-25n2-25n+100 = 0n-20n-5 = 0 n =20 or n = 5

Page No 19.31:

Question 29:

In an A.P. the first term is 2 and the sum of the first five terms is one fourth of the next five terms. Show that 20th term is −112.

Answer:

Given: a = 2, S5 = 14S10- S5We have: S5 =52 2×2+(5-1)dS5 = 52+2d   ....(i)Also, S10 = 1022×2+(10-1)dS10 = 54+9d  .....(ii) S5 = 14S10- S5 From (i) and (ii), we have: 52+2d = 14 5(4+9d)-5(2+2d)8+8d = 4+9d - 2-2dd = -6 a20 = a+20-1da20 = a+19d a20 = 2+19-6 a20 = -112

Page No 19.31:

Question 30:

If S1 be the sum of (2n + 1) terms of an A.P. and S2 be the sum of its odd terms, the prove that:
S1 : S2 = (2n + 1) : (n + 1)

Answer:

Let the A.P. be a, a+d, a+2d... S1 = 2n+122a+(2n+1-1)dS1 = 2n+122a+(2n)dS1 = (2n+1)(a+nd)           ...(i)S2 = n+122a+(n+1-1)×2dS2 = n+122a+2ndS2 = (n+1)a+nd             ...(ii)From (i) and (ii), we get:S1S2  = 2n+1n+1Hence, proved.

Page No 19.31:

Question 31:

Find an A.P. in which the sum of any number of terms is always three times the squared number of these terms.

Answer:

Given:
Sn = 3n2For n=1, S1 = 3×12 = 3For n=2, S2 = 3×22 = 12For n=3, S3 = 3×32 = 27    and so on S1 =a1 = 3a2 = S2-S1 = 12-3 = 9a3 = S3-S2 = 27-12 = 15and so onThus, the A.P. is  3, 9, 15...

Page No 19.31:

Question 32:

If the sum of n terms of an A.P. is nP + 12n (n − 1) Q, where P and Q are constants, find the common difference.

Answer:

We have:Sn = nP+12n(n-1)QFor n =1, S1 = P+0=PFor n =2, S2 = 2P+Q  Also, a1 = S1 = P,a2 = S2-S1 =2P+Q  -P = P+Q d = a2- a1 = P+Q-P = Q 

Page No 19.31:

Question 33:

The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.

Answer:

Let there be two A.P.s.Let their first terms be a1 and a2 and their common differences be d1 and d2.Given: 5n+49n+6 = Sum of n terms in the first A.P.Sum of n terms in the second A.P.5n+49n+6 = 2a1+[(n-1)d1]2a2+[(n-1)d2]Putting n=2×18-1= 35 in the above equation, we get:     5×35+49×35+6 = 2a1+34d12a2+34d2179321= a1+17d1a1+17d1179321= 18th term of the first A.P.18th term of the second A.P.

Page No 19.31:

Question 34:

The sums of first n terms of two A.P.'s are in the ratio (7n + 2) : (n + 4). Find the ratio of their 5th terms.

Answer:

Let the first term, the common difference and the sum of the first n terms of the first A.P. be a1, d1 and S1, respectively, and those of the second A.P. be a2, d2 and S2, respectively.Then, we have,  S1 = n22a1+n-1d1 And, S2 = n22a2+n-1d2Given: S1S2 = n22a1+n-1d1 n22a2+n-1d2 = 7n+2n+4S1S2 = 2a1+n-1d12a2+n-1d2= 7n+2n+4To find the ratio of the 5th terms of the two A.P.s, we replace n by (2×5-1 = 9) in the above equation: 2a1+9-1d12a2+9-1d2= 7×9+29+42a1+8d12a2+8d2= 7×9+29+4=6513 a1+4d1a2+4d2= 51=5:1



Page No 19.4:

Question 1:

If the nth term an of a sequence is given by an = n2n + 1, write down its first five terms.

Answer:

Given : an = n2-n+1For n=1, a1 = 12-1+1                         = 1For n=2, a2 = 22-2+1                         = 3For n=3, a3 = 32-3+1                         = 7For n=4, a4 = 42-4+1                         =13For n=5, a5 = 52-5+1                         = 21
Thus, the first five terms of the sequence are 1, 3, 7, 13, 21.

Page No 19.4:

Question 2:

A sequence is defined by an = n3 − 6n2 + 11n − 6, n ϵ N. Show that the first three terms of the sequence are zero and all other terms are positive.

Answer:

Given:
an = n3 − 6n2 + 11n − 6, n ϵ N
For n =1, a1 = 13-6×12+11×1-6 = 0For n = 2, a2 =  23-6×22+11×2-6 = 0For n =3, a3 = 33-6×32+11×3-6  = 0For n =4, a4 = 43-6×42+11×4-6  =6>0For n =5, a5 = 53-6×52+11×5-6 =24 >0and so onThus, the first three terms are zero and the rest of the terms are positive in the sequence.

Page No 19.4:

Question 3:

Let < an > be a sequence defined by
a1 = 3
and, an = 3an − 1 + 2, for all n > 1
Find the first four terms of the sequence.

Answer:

Given:
a1 = 3
And, an = 3an − 1 + 2 for all n > 1

a2 = 3a2-1+2 = 3a1+2 = 11a3 =  3a3-1+2 = 3a2+2 = 35a4 =  3a4-1+2 = 3a3+2 = 107Thus, the first four terms of the sequence are 3, 11, 35, 107.

Page No 19.4:

Question 4:

Let < an > be a sequence. Write the first five terms in each of the following:
(i) a1 = 1, an = an − 1 + 2, n ≥ 2
(ii) a1 = 1 = a2, an = an − 1 + an − 2, n > 2
(iii) a1 = a2 = 2, an = an − 1 − 1, n > 2

Answer:

(i) a1 = 1, an = an − 1 + 2, n ≥ 2
a2 = a1+2 = 1+2 = 3a3 = a2+2 = 5a4 = a3+2 = 7a5 = a4+2 = 9

Hence, the five terms are 1, 3, 5, 7 and 9.

(ii) a1 = 1 = a2, an = an − 1 + an − 2, n > 2
a3 =a2+a1 = 1+1=2a4 = a3+a2 =2+1 = 3a5 = a4+a3 = 3+2 = 5

Hence, the five terms are 1, 1, 2, 3 and 5.

(iii) a1 = a2 = 2, an = an − 1 − 1, n > 2
a3 = a2-1 = 2-1=  1a4 = a3-1 = 1-1=  0a5 = a4-1 = 0-1= -1

Hence, the five terms are 2, 2, 1, 0 and -1.

Page No 19.4:

Question 5:

The Fibonacci sequence is defined by
a1 = 1 = a2, an = an − 1 + an − 2 for n > 2

Find na+1na for n = 1, 2, 3, 4, 5.

Answer:

a1 = 1 = a2, an = an − 1 + an − 2 for n > 2
Then, we have:
a3 = a2+a1 = 1+1 = 2a4 = a3+a2 = 2+1 = 3a5  =a4+a3 = 3+2 = 5a6 = a5+a4 = 5+3 = 8For n=1, an+1an = a2a1 = 11=1For n=2, an+1an = a3a2 = 21=2For n=3, an+1an = a4a3 = 32For n=4, an+1an = a5a4 = 53For n=5, an+1an = a6a5 = 85

Page No 19.4:

Question 6:

Show that each of the following sequences is an A.P. Also find the common difference and write 3 more terms in each case.
(i) 3, −1, −5, −9 ...
(ii) −1, 1/4, 3/2, 11/4, ...
(iii) 2, 32, 52, 72, ...
(iv) 9, 7, 5, 3, ...

Answer:

(i) We have: -1-3 = -4,-5-(-1) = -4,-9-(-5) = -4...Thus, the sequence is an A.P. with the common difference being -4.The next three terms are as follows:-9-4 = -13-13-4 = -17-17-4 = -21(ii) We have:1/4-(-1) = 5/43/2-1/4 = 5/411/4-3/2 = 5/4Thus, the sequence is an A.P. with the common difference being (5/4).The next three terms are as follows:11/4+5/4 = 16/4 = 416/4+5/4 = 21/421/4+5/4 = 26/4

(iii) We have: 32 - 2 = 2252 - 32 = 2272 - 52 = 22Thus, the sequence is an A.P. with the common difference being (22).The next three terms are as follows:72+22 = 9292+22 = 112112+22 = 132(iv) We have: 7-9 = -25-7 = -23-5 = -2Thus, the sequence is an A.P. with the common difference being (-2).The next three terms are as follows:3-2 = 11-2 = -1-1-2 = -3

Page No 19.4:

Question 7:

The nth term of a sequence is given by an = 2n + 7. Show that it is an A.P. Also, find its 7th term.

Answer:

an = 2n+7 a1 = 2×1+7 = 9 a2 =2×2+7 =11 a3 =2×3+7 = 13a4 = 2×4+7 = 15and so onSo, common differenced=11-9=2Thus, the above sequence is an A.P. with the common difference as 2a7 = 2×7+7 = 21

Page No 19.4:

Question 8:

The nth term of a sequence is given by an = 2n2 + n + 1. Show that it is not an A.P.

Answer:

We have:
an= 2n2 + n + 1
a1 = 2×12+1+1      = 4,a2 = 2×22+2+1     = 11a3 = 2×32+3+1       = 22      a2-a1= 11-4       = 7 and a3-a2= 22-11                      = 11Since, a2-a1a3-a2Hence,it is not an AP.



Page No 19.42:

Question 1:

If 1a,1b,1c are in A.P., prove that:
(i) b+ca,c+ab,a+bc are in A.P.
(ii) a (b +c), b (c + a), c (a +b) are in A.P.

Answer:

Given: 1a,1b,1c are in A.P. 2b = 1a+1c2ac = ab+bc           ....(1)(i) To prove: b+ca, c+ab,a+bc are in A.P.      2a+cb = b+ca+a+bc2ac(a+c) = bc(b+c) +ab(a+b)LHS:  2ac(a+c)=(ab+bc)(a+c)          (From(1))= a2b+2abc+bc2 RHS: bc(b+c) +ab(a+b)= b2c+bc2 +a2b+ab2= b2c+ab2+bc2 +a2b=b(bc+ab)+bc2+a2b= 2abc+bc2+a2b =a2b+2abc+bc2         (From(1)) LHS= RHSHence, proved. 

(ii) To prove: a(b+c), b(c+a), c(a+b) are in A.P.2b(c+a) =a(b+c)+c(a+b)LHS: 2b(c+a)= 2bc+2baRHS: a(b+c)+c(a+b)= ab+ac+ac+bc= ab+2ac+bc=ab+ab+bc+bc         (From(1))= 2ab+2bcLHS = RHSHence, proved.

Page No 19.42:

Question 2:

If a2, b2, c2 are in A.P., prove that ab+c,bc+a,ca+b are in A.P.

Answer:

a2, b2, c2 are in A.P. 2b2 = a2+c2b2 -a2 = c2-b2(b+a)(b-a) = (c-b)(c+b)b-ac+b = c-bb+ab-a(c+a)(c+b)= c-b(b+a)(c+a)     Multiplying both the sides by 1c+a1c+a-1b+c = 1a+b-1c+a'  1b+c,1c+a,1a+b are in A.P.Multiplying each term by (a+b+c):a+b+cb+c, a+b+cc+a, a+b+ca+b are in A.P.Thus, ab+c+1 , bc+a+1 ,ca+b+1  are in A.P.Hence, ab+c, bc+a, ca+b are in A.P.

Page No 19.42:

Question 3:

If a, b, c are in A.P., then show that:
(i) a2 (b + c), b2 (c + a), c2 (a + b) are also in A.P.
(ii) b + ca, c + ab, a + bc are in A.P.
(iii) bca2, cab2, abc2 are in A.P.

Answer:

Since a, b, c are in A.P., we have:2b = a+c(i) We have to prove the following:2b2(a+c) = a2(b+c)+c2(a+b)LHS: 2b2 × 2b    (Given)     = 4b3RHS: a2b+a2c+ac2+c2b= ac(a+c) +b(a2+c2)= ac(a+c) +b[(a+c)2-2ac]=ac(2b)+b2b2-2ac=2abc +4b3-2abc=4b3LHS = RHS Hence, proved.    

(ii) We have to prove the following:2(c+a-b)=(b+c-a)+(a+b-c)LHS: 2(c+a-b)=2(2b-b)              2b=a+c=2bRHS: (b+c-a)+(a+b-c)=2bLHS=RHSHence, proved. 

(iii) We have to prove the following:2(ca-b2)=bc-a2+ab-c2RHS: bc-a2+ab-c2=c(b-c)+a(b-a)=ca+c2-c+aa+c2-a             2b=a+c=ca+c-2c2+aa+c-2a2=ca-c2+ac-a2=ca2-c22+ac2-a22=ac-12c2+a2=ac-124b2-2ac                a2+c2+2ac=4b2a2+c2=4b2 -2ac=ac-2b2+ac=2ac-2b2=2ac-b2=LHSHence, proved.

Page No 19.42:

Question 4:

If b+ca,c+ab,a+bc are in A.P., prove that:

(i) 1a,1b,1c are in A.P.

(ii) bc, ca, ab are in A.P.

Answer:

(i) Since b+ca,c+ab,a+bc are in A.P., we have:

    c+ab - b+ca = a+bc - c+abac+a2-b2-bcab = ab+b2-c2-acbca+ba-b+ca-bab = b+cb-c+ab-cbca-ba+b+cab = b-ca+b+cbca-bab = b-cbc1b-1a = 1c-1b

Hence, 1a,1b,1c are in A.P.


(ii) Since 1a,1b,1c are in A.P., we have:

1b-1a=1c-1ba-bab=b-cbca-ba=b-cca-bc = ab-cac - bc = ab - ac

Hence, bc, ca, ab are in A.P.

Page No 19.42:

Question 5:

If a, b, c are in A.P., prove that:
(i) (ac)2 = 4 (ab) (bc)
(ii) a2 + c2 + 4ac = 2 (ab + bc + ca)
(iii) a3 + c3 + 6abc = 8b3.

Answer:

Since a, b, c are in A.P., we have:
    2b = a+c
b = a+c2

(i) Consider RHS:
4 (ab) (bc)
Substituting b=a+c2: 4a-a+c2a+c2-c42a-a-c2a+c-2c2a-ca-ca-c2
Hence, proved.


(ii) Consider RHS:
2 (ab + bc + ca)
Substituting b=a+c2:2aa+c2+ca+c2+ac2a2+ac+ac+c2+2ac2a2+4ac+c2Hence, proved.


(iii) Consider RHS:
8b3
Substituting b=a+c2:8a+c23a3+c3+3aca+ca3+c3+3ac(2b)a3+c3+6abc

Hence, proved.

Page No 19.42:

Question 6:

If a1b+1c, b1c+1a, c1a+1b are in A.P., prove that a, b, c are in A.P.

Answer:

Given:
a1b+1c, b1c+1a, c1a+1b are in A.P.
By adding 1 to each term, we get:     a1b+1c+1, b1c+1a+1, c1a+1b+1 are in A.P. a1b+1c+1a, b1c+1a+1b, c1a+1b+1c are in A.P.Dividing all terms by 1a+1b+1c, we get:a, b, c are in A.P.        Hence, proved.

Page No 19.42:

Question 7:

Show that x2 + xy + y2, z2 + zx + x2 and y2 + yz + z2 are consecutive terms of an A.P., if x, y and z are in A.P.             [NCERT EXEMPLAR]

Answer:

As, x, y and z are in A.P.So, y=x+z2           .....iNow,x2+xy+y2+y2+yz+z2=x2+z2+2y2+xy+yz=x2+z2+2y2+yx+z=x2+z2+2x+z22+x+z2x+z                 Using i=x2+z2+2x+z24+x+z22=x2+z2+x+z22+x+z22=x2+z2+x+z2=x2+z2+x2+2xy+z2=2x2+2xy+2z2=2x2+xy+z2Since, x2+xy+y2+y2+yz+z2=2x2+xy+z2So, x2+xy+y2, x2+xy+z2 and y2+yz+z2 are in A.P.

Hence, x2 + xy + y2z2 + zx + x2 and y2 + yz + z2 are consecutive terms of an A.P.



Page No 19.46:

Question 1:

Find the A.M. between:
(i) 7 and 13
(ii) 12 and −8
(iii) (xy) and (x + y).

Answer:

(i) Let A1 be the A.M. between 7 and 13.

 A1 = a+b2

     = 7+132

     = 10

(ii) Let A1 be the A.M. between 12 and −8.

A1 = a+b2
    = 12+-82
    = 2

(iii) Let A1 be the A.M. between (xy) and (x + y).
A1 = a+b2
    = (x-y)+(x+y)2
    =  x

Page No 19.46:

Question 2:

Insert 4 A.M.s between 4 and 19.

Answer:

Let A1, A2, A3, A4 be the four A.M.s between 4 and 19. Then, 4, A1, A2, A3, A4 and 19 are in A.P. whose common difference is as follows:
d = 19-44+1
   = 3

A1=4+d =4+3=7A2=4+2d=4+6=10A3=4+3d=4+9=13A4=4+4d=4+12=16
Hence, the required A.M.s are 7, 10, 13, 16.

Page No 19.46:

Question 3:

Insert 7 A.M.s between 2 and 17.

Answer:

Let A1, A2, A3, A4, A5, A6, A7 be the seven A.M.s between 2 and 17.
Then, 2, A1, A2, A3, A4, A5, A6, A7  and 17 are in A.P. whose common difference is as follows:

d = 17-27+1

   =158

A1=2+d =2+158=318A2=2+2d =2+154=234A3=2+3d =2+458=618A4=2+4d =2+152=192A5=2+5d =2+758=918A6=2+6d =2+454=534A7=2+7d =2+1058=1218

Hence, the required A.M.s are 318, 234, 618, 192, 918, 534, 1218.

Page No 19.46:

Question 4:

Insert six A.M.s between 15 and −13.

Answer:

Let A1, A2, A3, A4, A5, A6 be the 6 A.M.s between 15 and -13.
Then, 15, A1, A2, A3, A4, A5, A6 and -13 are in A.P. whose common difference is as follows:

d = -13-156+1
   = -4

A1=15+d =15+-4=11A2=15+2d =15+-8=7A3=15+3d =15+-12=3A4=15+4d =15+-16=-1A5=15+5d =15+-20=-5A6=15+6d =15+-24=-9
Hence, the required A.M.s are 11, 7, 3, -1, -5, -9.

Page No 19.46:

Question 5:

There are n A.M.s between 3 and 17. The ratio of the last mean to the first mean is 3 : 1. Find the value of n.

Answer:

Let A1, A2, A3, A4 ....An be the n arithmetic means between 3 and 17.
Let d be the common difference of the A.P. 3, A1, A2, A3, A4....An and17.
Then, we have:
d = 17-3n+1 = 14n+1

Now, A1= 3 + d = 3 + 14n+1 = 3n+17n+1

And, An = 3+nd = 3+n14n+1 = 17n+3n+1

 AnA1=3117n+3n+13n+17n+1=3117n+33n+17=3117n+3=9n+518n=48n=6

Page No 19.46:

Question 6:

Insert A.M.s between 7 and 71 in such a way that the 5th A.M. is 27. Find the number of A.M.s.

Answer:

Let  A1, A2, A3, A4, 27, A6...An be the n arithmetic means between 7 and 71.
Thus, there are n+2 terms in all.
Let d be the common difference of the above A.P.
Now, a6 = 27
a+6-1d=27a+5d=27
d = 4

Also, 71 = an+2
71 = 7+n+2-14
71 = 7+n+14
n = 15

Therefore, there are 15 A.M.s.

Page No 19.46:

Question 7:

If n A.M.s are inserted between two numbers, prove that the sum of the means equidistant from the beginning and the end is constant.

Answer:

Let A1, A2......An be n A.M.s between two numbers a and b. Then, a, A1, A2.......An, b are in A.P. with common difference, d =b-an+1. A1+A2 +......+An=n2A1+An                                          =n2A1-d+An+d                                          =n2a+b                                          =n×a+b2                                          =A.M. between a and b, which is constant. 



Page No 19.47:

Question 8:

If x, y, z are in A.P. and A1 is the A.M. of x and y and A2 is the A.M. of y and z, then prove that the A.M. of A1 and A2 is y.

Answer:

x, y, z are in A.P.
y=x+z2

Now, A1 is the arithmetic mean of x and y.
A1=x+y2=x+x+z22=3x+z4

And, A2 is the arithmetic mean of y and z.
A2=y+z2=x+z2+z2=3z+x4

Let A3 be the arithmetic mean of A1 and A2.

A3=A1+A22     =3x+z4+3z+x42     =4x+4z8     =x+z2     =y

Hence, proved.

Page No 19.47:

Question 9:

Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Answer:

Let A1, A2, A3, A4, A5 be five numbers between 8 and 26.
Let d be the common difference.
Then, we have:
     26 = a7
26 = 8 + 7-1d
26 = 8 + 6d
d = 3

A1=8+d =8+3=11A2=8+2d=8+6=14A3=8+3d=8+9=17A4=8+4d=8+12=20A5=8+5d=8+15=23

Therefore, the five numbers are 11, 14, 17, 20, 23.



Page No 19.48:

Question 6:

A man is employed to count Rs 10710. He counts at the rate of Rs 180 per minute for half an hour. After this he counts at the rate of Rs 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.

Answer:

It is given that the man counts Rs 180 per minute for half an hour.
∴ Sum of money the man counts in 30 minutes = Rs 180×30 = Rs 5400

Total money counted by the man = Rs 10710
∴ Money left for counting after 30 minutes = Rs (10710 − 5400) = Rs 5310

It is given that after 30 minutes, he counts at the rate of Rs 3 less every minute than the preceding minute.

Therefore, it would be an A.P. where a = 177 and d = −3.
Let the time taken to count Rs 5310 be n minutes.
5310=n22×177+n-1×-310620=354 n-3n2+3n3n2-357n+10620=0n2-119n+3540=0n2-59n-60n+3540=0nn-59-60n-59=0n-59n-60=0 n=59 or 60

Thus, the time taken to count Rs 5310 would be 59 minutes or 60 minutes.

Hence, the total time taken to count Rs 10710 would be (30 + 59) minutes or (30 + 60) minutes, i.e. 89 minutes or 90 minutes, respectively.



Page No 19.49:

Question 1:

A man saved Rs 16500 in ten years. In each year after the first he saved Rs 100 more than he did in the receding year. How much did he save in the first year?

Answer:

Let the amount saved by the man in the first year be Rs A.
Let d be the common difference.
Let S10 denote the amount he saves in ten years.
Here, n =10, d =100

We know:
Sn=n22A+n-1d S10=1022A+10-110016500 = 52A+9003300 = 2A+900A = 1200

Therefore, the man saved Rs 1200 in the first year.

Page No 19.49:

Question 2:

A man saves Rs 32 during the first year. Rs 36 in the second year and in this way he increases his savings by Rs 4 every year. Find in what time his saving will be Rs 200.

Answer:

Let n be the time in which the man saved Rs 200.
Here, d= 4, a = 32
We know:
Sn=n22a+n-1d200 = n22×32+n-14400 = 64n+4n2-4n4n2+60n-400=0n2+15n-100=0n2+20n-5n-100=0n+20n-5=0n=5, n=-20            Rejecting the negative value

Therefore, the man took 5 years to save Rs 200.

Page No 19.49:

Question 3:

A man arranges to pay off a debt of Rs 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of the first instalment.

Answer:

Let S40 denote the total loan amount to be paid in 40 annual instalments.
S40 = 3600
Let Rs a be the value of the first instalment and Rs d be the common difference.
We know:

Sn=n22a+n-1d

4022a+40-1d = 3600202a+39d = 36002a+39d = 180                 ......1

Also, S40-S30=13×36003600 - S30=1200S30=24003022a+30-1d=2400152a+29d=24002a+29d=160                .....2

On solving equations 1 and 2, we get:
d = 2 and a =51

Hence, the value of the first instalment is Rs 51

Page No 19.49:

Question 4:

A manufacturer of radio sets produced 600 units in the third year and 700 units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find (i) the production in the first year (ii) the total product in 7 years and (iii) the product in the 10th year.

Answer:

Let an denote the production of radio sets in the nth year.
Here, a3 = 600, a7 = 700
We know:
an=a+n-1d

a3=a+2d600 = a+2d       .....1

And, a7=a+6d700=a+6d        .....2

Solving 1 and 2, we get:
d = 25, a = 550
Hence, the production in the first year is 550 units.

(ii)  Let Sn denote the total production in n years.
Total production in 7 years = S7
=722×550 + 7-125=4375 units

(iii)  Production in the 10th year = a10
  a10=a+10-1d      =550+925      =775

Page No 19.49:

Question 5:

There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.

Answer:

Let Sn be the total distance travelled by the gardener.
Let d be the common difference (distance) between two trees. Let a be the distance of the well from the first tree.
Here, n = 25, d = 10, a = 20
Distance travelled by the gardener from the well to the last tree = S25
S25=2522×20 + 25-110      =25240+240      =3500 m

Therefore, the total distance the gardener has to travel is 3500 m.

Page No 19.49:

Question 7:

A piece of equipment cost a certain factory Rs 600,000. If it depreciates in value, 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?

Answer:

Cost of the piece of equipment at the end of the first year = Rs 600000 – 15% of 600000
                                                                                      = Rs 600000 – Rs 90000
                                                                                      = Rs 510000
Cost of the piece of equipment at the end of the second year = Rs 600000 – 13.5% of 600000
                                                                                      = Rs 600000 – Rs 81000
                                                                                      = Rs 519000
Cost of the piece of equipment at the end of the third year = Rs 600000 – 12% of Rs 600000
                                                                                      = Rs 600000 – Rs 72000
                                                                                      = Rs 528000
The numbers 510000, 519000, 528000 are in an A.P. whose first term in 510000 and common difference is 9000.
∴ Cost of the piece of equipment at the end of 10 years = a + (10 – 1)d
                                                                                      = 510000 + 81000
                                                                                      = Rs 591000

Page No 19.49:

Question 8:

A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much the tractor cost him?

Answer:

Cost of the tractor = Rs 12000
It is given that the farmer pays Rs 6000 in cash.
Unpaid amount = Rs 6000
He has to pay Rs 6000 in annual instalments of Rs 500 plus 12% interest on the unpaid amount.
∴ Number of years taken by the farmer to pay the whole amount = 6000 ÷ 500 = 12
Hence, the interest paid by farmer annually would be as follows:
    12% of Rs 6000+12% of Rs 5500+12% of Rs 5000 ..=720+660+600....

It is in an A.P. where a = 720 , d = -60 and n = 12.
Total sum:
   1222×720+11×-60=61440-660=Rs 4680

∴ Amount the farmer has to pay = Rs 12000 + Rs 4680 = Rs 16680






                                                                  

Page No 19.49:

Question 9:

Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalments of Rs 1000 plus 10% interest on the unpaid amount. How much the scooter will cost him.

Answer:

Cost of the scooter = Rs 22000
Shamshad Ali pays Rs 4000 in cash.
∴ Unpaid amount = Rs 22000 - Rs 4000 = Rs 18000
Number of years taken by Shamshed Ali to pay the whole amount = 18000 ÷ 1000 = 18
He agrees to pay the balance in annual instalments of Rs 1000 plus 10% interest on the unpaid amount.

Total amount of instalments:
    10% of Rs 18000+10% of Rs 17000+10% of Rs 16000....=1800+1700+1600....

It is in an A.P. where a = 1800, d = -100 and n = 18.
Therefore, total amount of instalments:
   1822×1800+(18-1)×-100=93600-1700=Rs 17100

∴ Total amount Shamshad Ali has to pay = Rs (22000 + 17100) = Rs 39100

Page No 19.49:

Question 10:

The income of a person is Rs 300,000 in the first year and he receives an increase of Rs 10000 to his income per year for the next 19 years. Find the total amount, he received in 20 years.

Answer:

Let Sn denote the total amount the person receives in n years.
Let d be the common increment in his income every year.
Let a denote the initial income of the person.
Here, a = 300,000, d = 10000, n = 20

Total amount at the end of 20 years:
S20=2022×300,000+(20-1)10,000      =79,00,000

Therefore, the total amount the person receives in 20 years is Rs 79,00,000.

Page No 19.49:

Question 11:

A man starts repaying a loan as first instalment of Rs 100 = 00. If he increases the instalments by Rs 5 every month, what amount he will pay in the 30th instalment?

Answer:

Let a30 be the amount a man repays in the 30th instalment.
Let d be the common increment in his instalment every month.
Let a be the initial repayment.
Here, a = 100, d = 5, n = 30
Amount to be repaid in the 30th instalment:
a30a+n-1d
     =100+29×5=245

Hence, the man repays Rs 245 in his 30th instalment.

Page No 19.49:

Question 12:

A carpenter was hired to build 192 window frames. The first day he made five frames and each day thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?                                                                       [NCERT EXEMPLAR]

Answer:

We have,S=192, a=5, d=2Now,Sn=192n22a+n-1d=192n22×5+n-1×2=192n210+2n-2=192n22n+8=192nn+4=192n2+4n=192n2-12n+16n-192=0nn-12+16n-12=0n-12n+16=0n-12=0 or n+16=0n=12 or n=-16 n cannot be negative. n=12

So, the carpenter takes 12 days to finish the job.



Page No 19.50:

Question 13:

We know that the sum of the interior angles of a triangle is 180°. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.                                            [NCERT EXEMPLAR]

Answer:

We know that,
the sum of the interior angles of a polygon with 3 sides, a1 = 180°,
the sum of the interior angles of a polygon with 4 sides, a2 = 360°,
the sum of the interior angles of a polygon with 5 sides, a3 = 540°,
.
.
.

As, a2-a1=360°-180°=180° and a3-a2=540°-360°=180°i.e. a2-a1=a3-a2So, a1, a2, a3, ... are in A.P.Also, a=180° and d=180°Since, the sum of the interior angles of a 3 sided polygon=aSo, the sum of the interior angles of a 21 sided polygon=a19Now,a19=a+19-1d=180°+18×180°=180°+3240°=3420°

So, the sum of the interior angles for a 21 sided polygon is 3420°.

Page No 19.50:

Question 14:

In a potato race 20 potatoes are placed in a line at intervals of 4 meters with the first potato 24 metres from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?
                                                                                                                                                                                           [NCERT EXEMPLAR]

Answer:

We have,

the distance travelled to bring the first potato, a1 = 2 × 24 = 48 m,
the distance travelled to bring the second potato, a2 = 2 × (24 + 4) = 56 m,
the distance travelled to bring the third potato, a3 = 2 × (24 + 4 + 4) = 64 m,
.
.
.

As, a2-a1=56-48=8 and a3-a2=64-56=8i.e. a2-a1=a3-a2So, a1, a2, a3, ... are in A.P.Also, a=48, d=8, n=20Now,S20=2022a+20-1d=102×48+19×8=10×96+152=10×248=2480

So, he would have to run 2480 m to bring back all the potatoes.

Page No 19.50:

Question 15:

A man accepts a position with an initial salary of ₹5200 per month. It is understood that he will receive an automatic increase of ₹320 in the very next month and each month thereafter.
(i) Find his salary for the tenth month.
(ii) What is his total earnings during the first year?

Answer:

We have,

the initial salary, a1₹5200,
the salary of the second month, a2 = ₹5200 + ₹320 = â‚¹5520,
the salary of the third month, a3 = â‚¹5520 + â‚¹320 = â‚¹5840,
.
.
.

As, a2-a1=5520-5200=320 and a3-a2=5840-5520=320i.e. a2-a1=a3-a2So, a1, a2, a3, ... are in A.P.Also, a=5200, d=320i a10=a+10-1d=5200+9×320=5200+2880=8080So, the salary of the man for the tenth month is 8,080.ii S12=1222a+12-1d=62×5200+11×320=610400+3520=6×13920=83520So, the total earnings of the man during the first year is 83,520.

Page No 19.50:

Question 16:

A man saved ₹66000 in 20 years. In each succeeding year after the first year he saved ₹200 more than what he saved in the previous year. How much did he save in the first year?

Answer:

As, in each succeeding year after the first year he saved â‚¹200 more than what he saved in the previous year.
So, the savings of each year are in A.P.

We have,
the total savings of the man in 20 years, S20₹66000 and
the difference of his savings in each succeeding year, d₹200

Let his savings in the first year be a.

Now,
S20=660002022a+20-1d=66000102a+19×200=660002a+3800=66000102a=6600-3800a=28002 a=1400

So, he saved ₹1400 in the first year.

Page No 19.50:

Question 17:

In a cricket team tournament 16 teams participated. A sum of ₹8000 is to be awarded among themselves as prize money. If the last place team is awarded ₹275 in prize money and the award increases by the same amount for successive finishing places, then how much amount will the first place team receive?

Answer:

We have,
the total sum of prize money to be awarded among 16 teams, S16 = ₹8000 and
the prize money awarded to the last place team i.e. a16 = ₹275

As, the award increases by the same amount for successive finishing places.
So, the prize money are in A.P.

Let the prize money awarded to the first team be a.

Now,

S16=8000162a+a16=80008a+275=8000a+275=80008a=1000-275 a=725

So, the amount which the first place team will recieve is â‚¹725.

Page No 19.50:

Question 1:

Write the common difference of an A.P. whose nth term is xn + y.

Answer:

We have: an=xn+y a1=x+ya2=2x+y
Common difference of an A.P., d = a2-a1
2x+y - x+yx

Page No 19.50:

Question 2:

Write the common difference of an A.P. the sum of whose first n terms is p2n2+Qn.

Answer:

Sum of the first n terms of an A.P. = p2n2+Qn
Sum of one term of an A.P. = S1
p212+Q1p2+Q

Sum of two terms of an A.P. = S2
p222+Q22p+2Q
Now, we have:
a1+a2=S2p2+Q+ a2 = 2p+2Qa2 = Q+32p

Common difference:
d=a2-a1  =Q+32p-Q+p2  =p

Page No 19.50:

Question 3:

If the sum of n terms of an AP is 2n2 + 3n, then write its nth term.

Answer:

Given:
Sn=2n2+3n
S1=212+31          =5S2 = 222+32      =14 a1+a2=14   5+a2 =14   a2 = 9

Common difference, d  = a2-a1
                                      = 9 - 5
                                      = 4
nth term = a +n-1d
               = 5+n-14
               = 4n+1

Page No 19.50:

Question 4:

If log 2, log (2x − 1) and log (2x + 3) are in A.P., write the value of x.

Answer:

The numbers log 2, log (2x − 1) and log (2x + 3) are in A.P.

log 2x-1-log 2 = log 2x+3-log 2x-1log 2x-12=log 2x+32x-12x-12=2x+32x-12x-12=22x+322x+1-2.2x=2.2x+622x-4.2x-5=0

Let 2x=y.y2-4y-5=0y-5y+1=0y=5 or y=-12x=5   or 2x=-1  not possibleTaking log on both the sides, we get: log 2x=log5xlog2=log5x=log 5 log 2=log2 5x=log2 5

Disclaimer: The question in the book has some error, so, this solution is not matching with the solution given in the book. This solution here is created according to the question given in the book.

Page No 19.50:

Question 5:

If the sums of n terms of two arithmetic progressions are in the ratio 2n + 5 : 3n + 4, then write the ratio of their mth terms.

Answer:

Given:

SnSn1=2n+53n+4n22a+n-1dn22b+n-1d1=2n+53n+42a+n-1d2b+n-1d1=2n+53n+4         ...1

Ratio of their m terms = ambm
To find the ratio of the mth terms, replace n by 2m-1 in equation (1).
  2a+2m-2d2b+2m-2d1=22m-1+532m-1+4a+m-1db+m-1d1=4m-2+36m-3+4ambm=4m+16m+1

Page No 19.50:

Question 6:

Write the sum of first n odd natural numbers.

Answer:

We need to find the sum of 1, 3, 5, 7... upto n terms.
Here, a = 1, d = 2

We know:
Sn=n22a+n-1d     =n22×1+n-12     =n2

Therefore, the sum of the first n odd numbers is n2.

Page No 19.50:

Question 7:

Write the sum of first n even natural numbers.

Answer:

We need to find the sum of 2, 4, 6, 8...upto n terms.
Here, a = 2, d = 2

We know:
Sn=n22a+n-1d     =n22×2+n-12     =nn+1

Therefore, the sum of the first n odd numbers is nn+1.

Page No 19.50:

Question 8:

Write the value of n for which nth terms of the A.P.s 3, 10, 17, ... and 63, 65, 67, .... are equal.

Answer:

For the first series, a = 3, d1= 7
For the second series, b = 63, d2= 2

Given:
an=bna+n-1d1=b+n-1d23+n-17=63+n-123+7n-7=63+2n-25n=65n=13

Hence, the 13th terms of both the series are the same.



Page No 19.51:

Question 9:

If 3+5+7+...+upto n terms5+8+11+.... upto 10 terms 7, then find the value of n.

Answer:

3+5+7+...+upto n terms5+8+11+.... upto 10 terms=7n22×3+n-121022×5+10-13=7n4+2n370=7n2+2n-1295=0n2+37n-35n-1295=0n+37n-35n=35, n=-37Rejecting the negative value, we get:n=35

Page No 19.51:

Question 10:

If mth term of an A.P. is n and nth term is m, then write its pth term.

Answer:

Given:

am=na+m-1d=n        ....1an=ma+(n-1)d=m          ....2

Solving equations 1 and 2, we get:
d = -1
a = n+m-1

pth term:
ap=a+p-1d     = n+m-1 +p-1-1     =n+m-p

Hence, the pth term is n + m -p.

Page No 19.51:

Question 11:

If the sums of n terms of two AP.'s are in the ratio (3n + 2) : (2n + 3), then find the ratio of their 12th terms.

Answer:

Let the first terms of the two A.P.'s be a and a'; and their common difference be d and d'.

Now,

SnSn'=3n+22n+3n22a+n-1dn22a'+n-1d'=3n+22n+32a+n-1d2a'+n-1d'=3n+22n+3Let n=232a+23-1d2a'+23-1d'=3×23+22×23+32a+22d2a'+22d'=69+246+32a+11d2a'+11d'=7149 a12a12'=7149

So, the ratio of their 12th terms is 71 : 49.

Page No 19.51:

Question 1:

If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is
(a) 87
(b) 88
(c) 89
(d) 90

Answer:

(c) 89

a7=34a+6d=34        .....1Also, a13=64a+12d=64        .....2

Solving equations 1 and 2, we get:

a = 4 and d = 5

a18 = a+17d           =4+175            =89

Page No 19.51:

Question 2:

If the sum of p terms of an A.P. is q and the sum of q terms is p, then the sum of p + q terms will be
(a) 0
(b) pq
(c) p + q
(d) − (p + q)

Answer:

(d) -p+q

Sp=qp22a+p-1d=q2ap+p-1pd=2q          .....1Sq=pq22a+q-1d=p2aq+q-1qd=2p          .....2Multiplying equation 1 by q and equation 2 by p and then solving, we get:d=-2p+qpqNow, Sp+q=p+q22a+p+q-1d          =p22a+p-1d+qd + q22a+q-1d+pd         =Sp+pqd2+Sq+pqd2         =p+q+pqd          =p+q-2p+qpqpq         =-(p+q)

Page No 19.51:

Question 3:

If the sum of n terms of an A.P. be 3 n2n and its common difference is 6, then its first term is
(a) 2
(b) 3
(c) 1
(d) 4

Answer:

(a) 2

Sn=3n2-nS1=312-1S1=2 a1= 2

Page No 19.51:

Question 4:

Sum of all two digit numbers which when divided by 4 yield unity as remainder is
(a) 1200
(b) 1210
(c) 1250
(d) none of these.

Answer:

(b) 1210

The given series is 13, 17, 21....97.

a1=13, a2=17, an=97d=a2-a1=7-3=4

an=97a+n-1d=9713+n-14=97n=22

Sum of the above series:
S22= 2222×13+22-14        =1126+84       =1210

Page No 19.51:

Question 5:

In n A.M.'s are introduced between 3 and 17 such that the ratio of the last mean to the first mean is 3 : 1, then the value of n is
(a) 6
(b) 8
(c) 4
(d) none of these.

Answer:

(a) 6
Let A1, A2, A3, A4 .... An be the n arithmetic means between 3 and 17.
Let d be the common difference of the A.P. 3, A1, A2, A3, A4, .... An and 17.
Then, we have:

d = 17-3n+1 = 14n+1

Now, A1= 3 + d = 3 +14n+1 = 3n+17n+1

And, An = 3+nd = 3+n14n+1 = 17n+3n+1

 AnA1=3117n+3n+13n+17n+1=3117n+33n+17=3117n+3=9n+518n=48n=6

Page No 19.51:

Question 6:

If Sn denotes the sum of first n terms of an A.P. < an > such that SmSn=m2n2, thenaman=
(a) 2 m+12 n+1

(b) 2 m-12 n-1

(c) m-1n-1

(d) m+1n+1

Answer:

(b) 2 m-12 n-1

SmSn=m2n2m22a+m-1dn22a+n-1d=m2n22a+m-1d2a+n-1d=mn 2an+ndm-nd=2am+nmd-md2an-2am-nd+md=02an-m-dn-m=02an-m=dn-md=2a          .....1


Ratio of aman=a+m-1da+n-1d aman=a+m-2aa+n-2a           From (1)              = a+2am-2aa+2an-2a                 =2am-a2an-a              = 2m-12n-1

Page No 19.51:

Question 7:

The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be
(a) 5
(b) 6
(c) 7
(d) 8

Answer:

(b) 6

a= 1, an=11, Sn= 36 an= 11a+n-1d= 111+n-1d= 11n-1d= 10              .....1Also, Sn= 36n22a+n-1d=36n2+10=72                     Using (1)n=6

Page No 19.51:

Question 8:

If the sum of n terms of an A.P., is 3 n2 + 5 n then which of its terms is 164?
(a) 26th
(b) 27th
(c) 28th
(d) none of these.

Answer:

(b) 27th

Sn=3n2+5nS1=312+51 = 8 a1=8S2=322+52 = 22 a1+a2=22 a2 = 14Common difference, d =14-8= 6Also, an=164a+n-1d=1648+n-16=164n=27

Page No 19.51:

Question 9:

If the sum of n terms of an A.P. is 2 n2 + 5 n, then its nth term is
(a) 4n − 3
(b) 3 n − 4
(c) 4 n + 3
(d) 3 n + 4

Answer:

(c) 4 n + 3

Sn=2n2+5nS1=2.12+5.1 = 7 a1=7Sn=2.22+5.2=18 a1+a2=18 a2=11Common difference, d =11-7= 4an=a+n-1d    =7+n-14    =4n+3

Page No 19.51:

Question 10:

If a1, a2, a3, .... an are in A.P. with common difference d, then the sum of the series sin d [cosec a1 cosec a2 + cosec a1 cosec a3 + .... + cosec an − 1 cosec an] is
(a) sec a1 − sec an
(b) cosec a1 − cosec an
(c) cot a1 − cot an
(d) tan a1 − tan an

Answer:

(c) cot a1 − cot an
We have:
sin d cosec a1 cosec a2+cosec a2 cosec a3+....+cosec an-1 cosec an =sin dsin a1 sin a2+sin dsin a2 sin a3+.....+sin dsin an-1 sin an=sin (a2-a1)sin a1 sin a2+ sin (a3-a2)sin a2 sin a3+....+sin (an-an-1)sin an-1 sin an=sin a2 cos a1-cos a2 sin a1sin a1 sin a2+sin a3 cos a2-cos a3 sin a2sin a1 sin a2+.....+sin a2 cos a1-cos a2 sin a1sin a1 sin a2=cot a1-cot a2 +cot a2-cot a3+.....+cot an-1-cot an=cot a1-cot an



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Question 11:

In the arithmetic progression whose common difference is non-zero, the sum of first 3 n terms is equal to the sum of next n terms. Then the ratio of the sum of the first 2 n terms to the next 2 n terms is
(a) 1/5
(b) 2/3
(c) 3/4
(d) none of these

Answer:

(a) 1/5

S3n=S4n-S3n2S3n=S4n2×3n22a+3n-1d = 4n22a+4n-1d32a+3n-1d = 22a+4n-1d6a+9nd-3d=4a+8nd-2d2a+nd-d=02a+n-1d=0             ....1

Required ratio: S2nS4n-S2n

S2nS4n-S2n = 2n22a+2n-1d4n22a+4n-1d-2n22a+2n-1d                     =nnd2n3nd-nnd                     =16-1                     =15

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Question 12:

If a1, a2, a3, .... an are in A.P. with common difference d, then the sum of the series sin d [sec a1 sec a2 + sec a2 sec a3 + .... + sec an − 1 sec an], is
(a) sec a1 − sec an
(b) cosec a1 − cosec an
(c) cot a1 − cot an
(d) tan an − tan a1

Answer:

(d) tan an − tan a1
We have:

sin d sec a1 sec a2+sec a2 sec a3+....+sec an-1sec an =sin dcos a1cos a2+sin dcos a2 cos a3+.....+sin dcos an-1 cos an=sin (a2-a1)cos a1cos a2+ sin (a3-a2)cos a2 cos a3+....+sin (an-an-1)cos an-1 cos an=sin a2 cos a1-cos a2 sin a1cosa1 cos a2+sin a3 cos a2-cos a3 sin a2cos a1 cos a2+.....+sin a2 cos a1-cos a2 sin a1cos a1 cos a2=tan a1-tan a2 +tan a2-tan a3+.....+tan an-1-tan  an=tan a1-tan  an

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Question 13:

If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are
(a) 5, 10, 15, 20
(b) 4, 10, 16, 22
(c) 3, 7, 11, 15
(d) none of these

Answer:

(a) 5, 10, 15, 20

Let the four numbers in A.P. be as follows:
a-2d, a-d, a, a+d
Their sum = 50 (Given)
a-2d+a-d +a+a+d = 502a-d=25                .....1Also, a+d = 4a-2da+d=4a-8d3d=a                 ......2

From equations 1 and 2, we get:
d  = 5, a = 15

Hence, the numbers are 15-10, 15-5, 15, 15+5, i.e. 5, 10, 15, 20.

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Question 14:

If n arithmetic means are inserted between 1 and 31 such that the ratio of the first mean and nth mean is 3 : 29, then the value of n is
(a) 10
(b) 12
(c) 13
(d) 14

Answer:

(d) 14

The given series is 1, . . . . . . . . . . . , 31
There are n A.M.s between 1 and 31: 1, A1, A2, A3, . . . . ., An, 31.

Common difference, d = 31-1n+1 = 30n+1

Here, we have:
A1An=3291+d1+nd=3291+30n+11+n×30n+1=329n+1+30n+1+30n=329n+3131n+1=32929n+899=93n+364n=896n=14

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Question 15:

Let Sn denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = Sn k Sn1 + Sn − 2 , then k =
(a) 1
(b) 2
(c) 3
(d) none of these

Answer:

(b) 2

Let the A.P. be a, a+d, a+2d, a+3d...
Given:
d=Sn-kSn-1+Sn-2
For n = 3, we have:

d=3a+3d-k2a+d+a4a+2d-k2a+d=022a+d=k2a+d2=k

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Question 16:

The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by l2-a2k-(l+a), then k =
(a) S
(b) 2S
(c) 3S
(d) none of these

Answer:

(b) 2S

Given:

S=n2l+al+a=2Sn

Also, d=l2-a2k-l+ad=l+al-ak-l+ad=n-1d×2Snk-2Snk-2Sn=n-12Snk=2Snn-1+1k=2S

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Question 17:

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then k =
(a) 1n

(b) n-1n

(c) n+12n

(d) n+1n

Answer:

(d) n+1n

Given:
Sum of the even natural numbers = k × Sum of the odd natural numbers
n22a+n-1d=k×n22a+n-1d2×2+n-12=k×2×1+n-124+n-122+n-12=kn+1n=k

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Question 18:

If the first, second and last term of an A.P are a, b and 2a respectively, then its sum is
(a) ab2 (b-a)

(b) abb-a

(c) 3 ab2 (b-a)

(d) none of these

Answer:

(c) 3 ab2 (b-a)

Let the A.P. be a, a+d, a+2d........a+nd.
Here, let d be the common difference and n be the total number of terms.

Given:
a1=a,a2=ba+d=bd=b-a         .....1an=2aa+n-1d=2an-1d=ad=an-1         .....2

From equations 1 and 2, we have:
an-1=b-aab-a+1=na+b-ab-a=nbb-a=n

Now, sum of n terms of an A.P.:
S = n2a+an    =n23a    =3ab2b-a

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Question 19:

If, S1 is the sum of an arithmetic progression of 'n' odd number of terms and S2 the sum of the terms of the series in odd places, then S1S2 =
(a) 2nn+1

(b) nn+1

(c) n+12n

(d) n+1n

Answer:

(a) 2nn+1


Let n be an odd number.
Given:

S1=Sum of odd number of terms     =n22a+n-1d            .....1Since n is odd, the number of odd places = n+12S2=Sum of the terms of a series in odd places     =n+1222a+n+12-12d     =n+142a+n-1d           .....2

From equations 1 and 2, we have:

S1S2=n22a+n-1dn+142a+n-1d        =2nn+1

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Question 20:

If in an A.P., Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to
(a) 12p3

(b) mn p

(c) P3

(d) (m + n) p2

Answer:

(c) p3

Given:
Sn=n2pn22a+n-1d=n2p2a+n-1d=2np2a=2np-n-1d           .....1Sm=m2pm22a+m-1d=m2p2a+m-1d=2mp2a=2mp-m-1d           .....2

From 1 and 2, we have:

     2np-n-1d=2mp-m-1d2pn-m=dn-1-m+12p=d

Substituting d = 2p in equation 1, we get:
a = p

Sum of p terms of the A.P. is given by:

p22a+p-1d=p22p+p-12p =p3

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Question 21:

Mark the correct alternative in the following question:

If in an A.P., the pth term is q and (p + q)th term is zero, then the qth term is

(a) -p                         (b) p                         (c) + q                         (d) p - q

Answer:

As, ap=qa+p-1d=q             .....iAlso, ap+q=0a+p+q-1d=0       .....iiSubtracting i from ii, we geta+p+q-1d-a-p-1d=0-qp+q-1-p+1d=-qqd=-qd=-qqd=-1Substituting d=-1 in i, we geta+p-1×-1=qa-p+1=qa=p+q-1Now,aq=a+q-1d=p+q-1+q-1×-1=p+q-1-q+1=p

Hence, the correct alternative is option (b).

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Question 22:

Mark the correct alternative in the following question:

The 10th common term between the A.P.s 3, 7, 11, 15, ... and 1, 6, 11, 16, ... is

(a) 191                         (b) 193                         (c) 211                         (d) none of these

Answer:

As, the common difference of the A.P. 3, 7, 11,15, ...=7-3=4 andthe common difference of the A.P. 1, 6, 11, 16, ...=6-1=5And, the common terms of both the A.P.s will be in A.P.So, the common difference of the A.P. of the common terms, d=LCM4, 5=4×5=20 andits first common term, a=11Now, the tenth common term, a10=a+10-1d=11+9×20=11+180=191

Hence, the correct alternative is option (a).

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Question 23:

Mark the correct alternative in the following question:

If in an A.P. Sn=n2q and Sm=m2q, where Sr denotes the sum of r terms of the A.P., then Sq equalsa q32                    b mnq                    c q3                    d m2+n2q

Answer:

As, Sn=n2qn22a+n-1d=n2q2a+n-1d=n2q×2n2a+n-1d=2nq         .....iAlso, Sm=m2qm22a+m-1d=m2q2a+m-1d=m2q×2m2a+m-1d=2mq         .....iiSubtracting i from ii, we get2a+n-1d-2a-m-1d=2nq-2mqn-1-m+1d=2qn-mn-md=2qn-md=2qn-mn-md=2qSubstituting d=2q in ii, we get2a+m-12q=2mq2a+2mq-2q=2mq2a=2qa=qNow,Sq=q22a+q-1d=q22q+q-12q=q22q+2q2-2q=q22q2=q3

Hence, the correct alternative is option (c).



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Question 24:

Mark the correct alternative in the following question:

Let Sn denote the sum of first n terms of an A.P. If S2n = 3Sn, then S3n : Sn is equal to

(a) 4                                  (b) 6                                  (c) 8                                  (d) 10

Answer:

As, S2n=3Sn2n22a+2n-1d=3n22a+n-1d22a+2n-1d=32a+n-1d4a+22n-1d=6a+3n-1d4a+4nd-2d=6a+3nd-3d6a-4a+3nd-3d-4nd+2d=02a-nd-d=02a-n+1d=02a=n+1d             .....iNow,S3nSn=3n22a+3n-1dn22a+n-1d=3n+1d+3n-1dn+1d+n-1d                  Using i=3nd+d+3nd-dnd+d+nd-d=34nd2nd=3×2=6

Hence, the correct alternative is option (b).



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