Rd Sharma Xi 2018 Solutions for Class 11 Humanities Math Chapter 20 Geometric Progressions are provided here with simple step-by-step explanations. These solutions for Geometric Progressions are extremely popular among Class 11 Humanities students for Math Geometric Progressions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2018 Book of Class 11 Humanities Math Chapter 20 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2018 Solutions. All Rd Sharma Xi 2018 Solutions for class Class 11 Humanities Math are prepared by experts and are 100% accurate.

Page No 20.10:

Question 2:

Show that the sequence <an>, defined by an = 23n, n ϵ N is a G.P.

Answer:

We have,an=23n, nNPutting n=1, 2, 3, ...a1=231=23, a2=232=29, a3=233=227 and so on.Now, a2a1=2923=13, a3a2=22729=13 and so on.a2a1=a3a2= ... =13So, the sequence is an G.P., where 23 is the first term and 13 is the common ratio.

Page No 20.10:

Question 3:

Find:
(i) the ninth term of the G.P. 1, 4, 16, 64, ...

(ii) the 10th term of the G.P. -34,12,-13,29, ...

(iii) the 8th term of the G.P. 0.3, 0.06, 0.012, ...

(iv) the 12th term of the G.P. 1a3 x3, ax, a5 x5 , ...

(v) nth term of the G.P. 3,13,133, ...

(vi) the 10th term of the G.P. 2,12,122, ...

Answer:

i Here, First term, a=1 Common ratio, r=a2a1=41=49th term=a9=ar(9-1)=1(4)8=48=65536Thus, the 9th term of the given GP is 65536.

iiHere, First term, a=-34 Common ratio, r =a2a1=12-34=-2310th term=a10=ar(10-1)=-34-239=12238Thus, the 10th term of the given GP is 12238.

iii Here, First term, a=0.3Common ratio, r =a2a1=0.060.3=0.28th term=a8=ar(8-1)=0.3(0.2)7Thus, the 8th term of the given GP is 0.3(0.2)7.

iv Here, First term, a=1a3x3Common ratio, r =a2a1=ax1a3x3=a4x412th term=a12=ar(12-1)=1a3x3(a4x4)11=a41x41Thus, the 12th term of the given GP is a41x41.


v Here, First term, a=3Common ratio, r =a2a1=133=13nth term=an=ar(n-1)=313n-1Thus, the nth term of the given GP is 313n-1.

viHere, First term, a=2Common ratio, r =a2a1=122=1210th term=a10=ar(10-1)=2129=12×128Thus, the 10th term of the given GP is 12×128.

Page No 20.10:

Question 4:

Find the 4th term from the end of the G.P. 227,29,23, ..., 162

Answer:

Here, first term, a=227Common ratio, r=a2a1=29227=3 Last term, l=162After reversing the given G.P., we get another G.P. whose first term is l and common ratio is 1r. 4th term from the end = l1r4-1=(162)133=6

Page No 20.10:

Question 5:

Which term of the progression 0.004, 0.02, 0.1, ... is 12.5?

Answer:

We have,a2a1=0.020.004=5, a3a2=0.10.02=5a2a1=a3a2=5The given progression is a G.P. whose first term, a is 0.004 and common ratio, r is 5.Let the nth term be 12.5. an=12.5 arn-1=12.5(0.004)(5)n-1=12.5(5)n-1=12.50.004(5)n-1=3125(5)n-1=(5)5Comparing the power of both the sidesn-1=5n=6Thus, 6th term of the given G.P. is 12.5

Page No 20.10:

Question 6:

Which term of the G.P. :

(i) 2,12,122,142, ... is15122?

(ii) 2, 22, 4, ... is 128 ?

(iii) 3, 3, 33, ... is 729 ?

(iv) 13,19,127 ...is 119683 ?

Answer:

i Here, first term, a=2  and common ratio, r=12Let the nth term be 15122.  an = 15122 arn-1 = 15122 212n-1 = 15122 12n-1 = 11024 12n-1 = 1 210 n-1 = 10 n=11Thus, the 11th term of the given G.P. is 15122.

ii Here, first term, a=2 and common ratio, r=2Let the nth term be 128.  an = 128 arn-1 = 128 22n-1 = 1282 (2)n-1 = 128 2n-1 = 64 2n-1 = 212 n-1 = 12 n = 13Thus, the 13th term of the given G.P. is 128.

iii Here, first term, a=3 and common ratio, r=3Let the nth term be 729.  an=729 arn-1 = 729 33n-1 = 729 (3)n-1 = 3123=(3)11n-1 = 11n=12Thus, the 12th term of the given G.P. is 729.

(iv) Here, first term, a=13 and common ratio, r=13Let the nth term be 119683.  an=119683 arn-1 = 119683 1313n-1 = 119683 13n-1 = 339=138 n-1 = 8 n=9Thus, the 9th term of the given G.P. is 119683.

Page No 20.10:

Question 7:

Which term of the progression 18, −12, 8, ... is 512729?

Answer:

Here, first term, a = 18 and common ratio, r = -23Let the nth term be 512729. arn-1 = 51272918-23n-1=512729-23n-1=512729×118 =2566561-23n-1=-238n-1 = 8 n=9Thus, the 9th term of the given G.P. is 512729.

Page No 20.10:

Question 8:

Find the 4th term from the end of the G.P. 12,16,118,154, ...,14374

Answer:

After reversing the given G.P., we get another G.P. whose first term, l is 14374 and common ratio is 3. 4th term from the end=l1r4-1 =14374 34-1=274374= 1162 

Page No 20.10:

Question 9:

The fourth term of a G.P. is 27 and the 7th term is 729, find the G.P.

Answer:

Let a be the first term and r be the common ratio of the given G.P.  a4 =27 and a7 = 729ar3 = 27 and ar6 = 729ar6ar3 = 72927 r3 = 33  r = 3Putting r = 3 in ar3 = 27a33 = 27 a = 1Thus, the given G.P. is 1, 3, 9, ...

Page No 20.10:

Question 10:

The seventh term of a G.P. is 8 times the fourth term and 5th term is 48. Find the G.P.

Answer:

Let a be the first term and r be the common ratio. a7=8a4  and a5=48ar6=8ar3 and ar4=48r3= 8 r3=23r = 2Putting r = 2 in ar4=48a24 = 48 a=3Thus, the given G.P. is 3, 6, 12, ...  

Page No 20.10:

Question 11:

If the G.P.'s 5, 10, 20, ... and 1280, 640, 320, ... have their nth terms equal, find the value of n.

Answer:

Given:First term, a=5 Common ratio, r=2an= 52n-1                                                               ... 1Similarly, an= 128012n-1                                   ... 2From 1 and 252n-1 = 128012n-11256=14n-1 144=14n-1    n-1 = 4 n = 5

Page No 20.10:

Question 12:

If 5th, 8th and 11th terms of a G.P. are p. q and s respectively, prove that q2 = ps.

Answer:

Let a be the first term and r be the common ratio of the given G.P. p = 5th term p = ar4                  ... 1q = 8th term q = ar7                    ... 2s = 11th  s = ar10                ... 3Now, q2 = ar72 = a2r14ar4 ar10 = ps           From 1 and 3  q2 = ps

Page No 20.10:

Question 13:

The 4th term of a G.P. is square of its second term, and the first term is − 3. Find its 7th term.

Answer:

Let r be the common ratio of the given G.P. Then, a4 = a22                   GivenNow, ar3 = a2r2  r = a     r = -3                             Putting a=-3  a7  = ar6   a7  = -3 -36             Putting a=-3 and r=-3  a7=-3-729  a7=-2187Thus, the 7th term of the G.P. is -2187.                

Page No 20.10:

Question 14:

In a GP the 3rd term is 24 and the 6th term is 192. Find the 10th term.

Answer:

Let a be the first term and r be the common ratio. a3 = 24 and a6 = 192ar2 = 24 and ar5 = 192ar5ar2=19224r3 = 8 r3 =23 r=2Putting r=2 in  ar2=24a22 = 24 a = 6Now, 10th term =a10 =ar9Putting a=6 and r=2 in a10 = ar9a10= 629 =3072Thus, the 10th term of the G.P. is 3072.    

Page No 20.10:

Question 15:

If a, b, c, d and p are different real numbers such that:
(a2 + b2 + c2) p2 − 2 (ab + bc + cd) p + (b2 + c2 + d2) ≤ 0, then show that a, b, c and d are in G.P.

Answer:

a2+b2+c2p2-2ab+bc+cdp+b2+c2+d20a2p2+b2p2+c2p2-2abp+bcp+cdp+b2+c2+d20a2p2-2abp+b2+b2p2-2bcp+c2+c2p2-2cdp+d20ap-b2+bp-c2+cp-d20ap-b2+bp-c2+cp-d2=0ap-b2=0 p = baAlso, bp-c2=0 p = cbSimiliarly, cp-d2=0 p = dc ba= cb=dcThus, a, b, c and d are in G.P.

Page No 20.10:

Question 16:

If a+bxa-bx=b+cxb-cx=c+dxc-dx (x ≠ 0), then show that a, b, c and d are in G.P.

Answer:

Given:a+bxa-bx=b+cxb-cx=c+dxc-dxNow, a+bxa-bx=b+cxb-cxApplying componendo and dividendoa+bx+a-bxa+bx-a-bx=b+cx+b-cxb+cx-b-cx2a2bx = 2b2cxab=bcSimiliarly, b+cx+b-cxb+cx-b-cx=c+dx+c-dxc+dx-c-dx bc = cdTherefore, a, b, c and d are in G.P. 

Page No 20.10:

Question 17:

If the pth and qth terms of a G.P. are q and p, respectively, then show that (p + q)th term is qppq1p-q.

Answer:

As, ap=qarp-1=q           .....iAlso, aq=parq-1=p           .....iiDividing i by ii, we getarp-1arq-1=qprp-1-q+1=qprp-q=qpr=qp1p-qSubstituting the value of r in ii, we getaqp1p-qq-1=paqpq-1p-q=pa=p×pqq-1p-qa=ppqq-1p-qNow,ap+q=arp+q-1=ppqq-1p-q×qp1p-qp+q-1=ppqq-1p-q×qpp+q-1p-q=pqp-q-1p-q×qpp+q-1p-q=p×qp-q-1p-q+p+q-1p-q=p×qp-q+1+p+q-1p-q=p×qppp-q=p×qpp-qppp-q=qpp-qppp-q-1=qpp-qpp-p+qp-q=qpp-qpqp-q=qp×1p-qpq×1p-q=qppq1p-q



Page No 20.16:

Question 1:

Find three numbers in G.P. whose sum is 65 and whose product is 3375.

Answer:

Let the terms of the the given G.P. be ar, a and ar.
Then, product of the G.P. = 3375
a3 = 3375
a = 15
Similarly, sum of the G.P. = 65
ar+a+ar = 65
Substituting the value of a
15r+15+15r=6515r2+15r+15=65r15r2-50r+15=053r2-10r+3=03r2-10r+3 = 03r-1r-3=0r=13, 3
Hence, the G.P. for a = 15 and r = 13 is 45, 15, 5.
And, the G.P. for a = 15 and r = 3 is 5, 15, 45.

Page No 20.16:

Question 2:

Find three numbers in G.P. whose sum is 38 and their product is 1728.

Answer:

Let the terms of the the given G.P. be ar, a and ar.
Then, product of the G.P. = 1728
a3 = 1728
a = 12
Similarly, sum of the G.P. = 38
ar+a+ar = 38
Substituting the value of a
12r+12+12r=3812r2+12r+12=38r12r2-26r+12=026r2-13r+6=06r2-13r+6 = 03r-22r-3=0r=23,32
Hence, the G.P. for a = 12 and r23 is 18, 12 and 8.
And, the G.P. for  a = 12 and r = 32is 8, 12 and 18.

Hence, the three numbers are 8, 12 and 18.

Page No 20.16:

Question 3:

The sum of first three terms of a G.P. is 13/12 and their product is − 1. Find the G.P.

Answer:

Let the first three numbers of the given G.P. be ar, a and ar.
∴ Product of the G.P. = −1
a3 = −1
a = −1
Similarly, Sum of the G.P. = 1312   
ar+a+ar = 1312
Substituting the value of a = −1
-1r-1-r=131212r2+25r+12=012r2+16r+9r+12=04r3r+4+33r+4=04r+33r+4 = 0r=-34,-43
Hence, the G.P. for a = −1 and r = -34 is 43,-1 and 34.

And, the G.P. for a = −1 and r = -43is 34,-1 and 43.

Page No 20.16:

Question 4:

The product of three numbers in G.P. is 125 and the sum of their products taken in pairs is 8712. Find them.

Answer:

Let the required numbers be ar, a and ar.
Product of the G.P. = 125
a3=125  a = 5
Sum of the products in pairs = 8712=1752

ar×a+a×ar+ar×ar=1752a2r+a2r+a2=1752Substituting the value of a25r +25r+25=175250r2+50r+50 =175r50r2-125r+50=025(2r2-5r+2)=02r2-4r-r+2=02r(r-2)-1(r-2)=0(2r-1)(r-2)=0 r = 12, 2

Hence, the G.P. for a = 5 and r = 12 is 10, 5 and 52.
And, the G.P. for a = 5 and r = 2 is 52, 5 and 10.

Page No 20.16:

Question 5:

The sum of first three terms of a G.P. is 3910 and their product is 1. Find the common ratio and the terms.

Answer:

Let the terms of the G.P be ar, a and ar.
∴ Product of the G.P. = 1
a3=1a=1
Now, sum of the G.P. = 3910
ar+a+ar=3910a1r+1+r=391011r+1+r=391010r2+10r+10=39r10r2-29r+10=010r2-25r-4r+10=05r(2r-5)-2(2r-5)=05r-22r-5=0r=25, 52
Hence, putting the values of a and r , the required numbers are 52, 1, 25 or 25, 1 and 52.

Page No 20.16:

Question 6:

The sum of three numbers in G.P. is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in A.P. Find the numbers.

Answer:

Let the numbers be a, ar and ar2.
Sum=14 a+ar+ar2=14                      a(1+r+r2)=14                   ... i
According to the question, a + 1, ar + 1 and ar2 − 1 are  in A.P.
 2ar+1 = a+1 + ar2-12ar + 2  = a + ar22ar + 2 = 14-ar                                    [Fromi]3ar = 12 a = 4r                                                    ... iiPutting a = 4r in i4r(1+r+r2)=144r2-10r+4 = 0 4r2-8r-2r+4 = 0 4r-2r-2=0r=12, 2
Putting r = 12in ii, we get a = 8.So, the G.P. is 8, 4 and 2.

Similarly putting r = 2 in (ii), we get a = 2.
So, the G.P is 2, 4 and 8.

Page No 20.16:

Question 7:

The product of three numbers in G.P. is 216. If 2, 8, 6 be added to them, the results are in A.P. Find the numbers.

Answer:

Let the terms of the given G.P. be ar, a and ar.
∴ Product = 216
a3=216a=6
It is given that ar+2, a+8 and ar+6 are in A.P.
  2a+8=ar+2+ar+6Putting a = 6, we get28 = 6r+2+6r+628r = 6r2+8r+66r2 -20r+6=06r-2r-3 =0r = 13, 3Hence, putting the values of a and r, the required numbers are 18, 6, 2 or 2, 6 and 18.

Page No 20.16:

Question 8:

Find three numbers in G.P. whose product is 729 and the sum of their products in pairs is 819.

Answer:

Let the required numbers be ar, a and ar.
Product of the G.P. = 729
a3=729a=9
Sum of the products in pairs = 819
ar×a+a×ar+ar×ar=819a21r+r+1 = 819811+r2+rr= 8199r2+r+1=91r9r2-82r+9=09r2-81r-r+9=09r-1r-9=0r=19, 9Hence, putting the values of a and r, we get the numbers to be  81, 9 and 1 or 1, 9 and 81.

Page No 20.16:

Question 9:

The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.

Answer:

Let the required numbers be a, ar and ar2.
Sum of the numbers = 21
a+ar+ar2=21a(1+r+r2)=21            ...(i)                  
Sum of the squares of the numbers = 189
a2+(ar)2+(ar2)2 = 189
a2+(ar)2+(ar2)2 = 189 a21+r2+r4 = 189                            ...(ii)                                     
Now, a ( 1 + r + r2) = 21                       [From (i)]Squaring both the sidesa21 + r + r22= 441a2 1 + r2+r4 + 2a2r1+r+r2 = 441189 + 2ara1+r+r2 = 441         [Using (ii)]189 +2ar×21 = 441                       [Using (i)]ar = 6 a=6r                                                 ...(iii)Putting a = 6r in (i) 6r1+r+r2=216r+6+6r=216r2 +6r+6=21r6r2-15r+6=03(2r2-5r+2)=02r2-5r+2 = 0(2r-1)(r-2)=0r = 12, 2Putting r = 12 in a=6r, we get a=12. So, the numbers are 12, 6 and 3.Putting r = 2 in a=6r, we get a=3. So, the numbers are 3, 6 and 12.Hence, the numbers that are in G.P are 3, 6 and 12.



Page No 20.27:

Question 1:

Find the sum of the following geometric progressions:
(i) 2, 6, 18, ... to 7 terms;
(ii) 1, 3, 9, 27, ... to 8 terms;
(iii) 1, −1/2, 1/4, −1/8, ... to 9 terms;
(iv) (a2b2), (ab), a-ba+b, ... to n terms;
(v) 4, 2, 1, 1/2 ... to 10 terms.

Answer:

(i) Here, a = 2 and r = 3.
 S7 = ar7-1r-1 =2 37-13-1 =2187-1=2186

(ii) Here, a = 1 and r = 3.
 S8 = ar8-1r-1 =1 38-13-1 =6561-12=3280

(iii) Here, a = 1 and r = −12.
 S9 = a1-r91-r = 1 1--1291--12 =1--151232=51351232=513×2512×3=171256

(iv) Here, a = a2b2 and r = 1a+b.
 Sn = a1-rn1-r = a2-b2 1-1a+bn1-1a+b =a2-b2a+bn-1a+bna+b-1a+bSn=a+ba-ba+bn-1a+bn-1a+b-1=a-ba+bn-2a+bn-1a+b-1

(v) Here, a = 4 and r = 12
 Sn=a1-r101-r=41-12101-12=41-1102412=81-11024=1023128

Page No 20.27:

Question 2:

Find the sum of the following geometric series:
(i) 0.15 + 0.015 + 0.0015 + ... to 8 terms;

(ii) 2+12+122+ ... to 8 terms;

(iii) 29-13+12-34+... to 5 terms;

(iv) (x +y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + ... to n terms;

(v) 35+452+353+454+... to 2n terms;

(vi) a1+i+a(1+i)2+a(1+i)3+...+a(1+i)n.

(vii) 1, −a, a2, −a3, ... to n terms (a ≠ 1)

(viii) x3, x5, x7, ... to n terms

(ix) 7, 21, 37, ... to n terms

Answer:

(i) Here, a = 0.15 and r =a2a1=0.0150.15=110.
S8=a1-r81-r =0.151-11081-110=0.151-1108110=161-1108

(ii) Here, a = 2 and r = 12.
S8=a1-r81-r=21-1281-12=21-125612=22255256=2552128

(iii) Here, a =29 and r = -32.
 S5=ar5-1r-1=29-325-1-32-1=29-24332-1-32-1=29-27532-52= 11001440=5572                      

             



Page No 20.28:

Question 3:

Evaluate the following:
(i) n=111(2+3n)

(ii) k=1n(2k+3k-1)

(iii) n=2104n

Answer:

(i)
S11 = n=1112+3nS11 = n=1112  + n=1113n S11 = 2 × 11 + 3+32+33+ ... +311= 22  +3311-13-1 =22 +177147-12= 22+265719 = 265741

(ii)
 Sn =k=1n2k+3k-1=k=1n2k   + k=1n3k-1=2 + 4 + 8+ ... +2n     + 1 + 3 + 9+ ... +3n  =22n-12-1 + 13n-13-1 = 122n+2-4+3n-1 = 122n+2+3n-5    

(iii)
n=2104n= 42 + 43+44+ ... +410=16 + 64 +256+ ... + 410=1649-14-1 = 16349-1      
       

Page No 20.28:

Question 4:

Find the sum of the following series:
(i) 5 + 55 + 555 + ... to n terms;
(ii) 7 + 77 + 777 + ... to n terms;
(iii) 9 + 99 + 999 + ... to n terms;
(iv) 0.5 + 0.55 + 0.555 + ... to n terms.
(v) 0.6 + 0.66 + 0.666 + .... to n terms

Answer:

(i) We have,
5 + 55 + 555+ ... n terms
Taking 5 as common:
Sn = 5[1 + 11 + 111 + ... n terms]
= 599+99+999+ ... n terms= 5910-1+102-1+103-1+ ... +10n-1= 5910+102+103+ ... +10n - 1+1+1+1+ ... n times= 5910 × 10n-110-1 - n = 59 10910n-1 - n= 58110n+1-9n-10

(ii) We have,
7 + 77 + 777 + ... n terms
Sn= 7 [1 + 11 + 111 + ... n terms]
= 799+99+999+ ...  n terms= 7910-1+102-1+103-1+ ... +10n-1= 7910+102+103+ ... +10n - 1+1+1+1 ... n times= 7910 × 10n-110-1 - n  = 79 10910n-1 - n= 78110n+1-9n-10

(iii) We have,
9 + 99 + 999 + ... n terms
= 9+99+999+ ... + to n terms= 10-1+102-1+103-1+ ... +10n-1= 10+102+103+ ... +10n - 1+1+1+1 ... n times= 10 × 10n-110-1 - n  = 10910n-1 - n= 1910n+1-9n-10

(iv) We have,
0.5 + 0.55 + 0.555 + ... n terms
Sn= 5 [0.1 + 0.11+0.111 + ... n terms]
= 590.9+0.99+0.999+ ... + to n terms= 59910+9100+91000+ ... n terms= 591-110+1-1100+1-11000+ ... n terms = 59n-110+1102+1103+ ... n terms = 59n-1101-110n1-110= 59n-191-110n
 
(v) We have,
 0.6 + 0.66 +.666 + ... to n terms
Sn= 6 [0.1 + 0.11+ 0.111 + ... n terms]
= 690.9+0.99+0.999+ ... n terms= 69910+9100+91000+ ... n terms= 691-110+1-1100+1-11000+ ... n terms = 69n-110+1102+1103+ ... n terms = 69n-1101-110n1-110= 69n-191-110n    
 

Page No 20.28:

Question 5:

How many terms of the G.P. 3, 3/2, 3/4, ... be taken together to make 3069512?

Answer:

Here, a = 3
Common ratio, r = 12
Sn = 3069512
 Sn= 31-12n1-123069512 = 31-12n12 3069512= 6 1-12n30693072=1-12n  12n= 1 - 30693072  12n= 330722n = 307232n=1024 2n= 210 n = 10

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Question 6:

How many terms of the series 2 + 6 + 18 + ... must be taken to make the sum equal to 728?

Answer:

Here, a = 2
Common ratio, r = 3
Sum of n terms, Sn = 728
Sn=23n-13-1 728 = 23n-12728 =3n-1     3n = 7293n= 36 n = 6   

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Question 7:

How many terms of the sequence 3, 3, 33, ... must be taken to make the sum 39+133?

Answer:

Here, a = 3
Common ratio, r3
Sum of n terms, Sn = 39+33

Sn=33n-13-1 39+133 = 33-13n-13n-1 =  39+1333-133n = 1 + 263n = 27 3n = 36 n = 6   

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Question 8:

The sum of n terms of the G.P. 3, 6, 12, ... is 381. Find the value of n.

Answer:

Here, a = 3
Common ratio, r = 3
Sum of n terms, Sn = 381
∴ Sn = 3 + 6 + 12 + ... + n terms
 381 = 32n-12-1 381 = 3 2n-1127 = 2n-12n=128   2n = 27  n = 7

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Question 9:

The common ratio of a G.P. is 3 and the last term is 486. If the sum of these terms be 728, find the first term.

Answer:

Here, common ratio, r = 3
nth term, an = 486
Sn = 728

an=486 arn-1=486a3n-1 = 486  a3n = 486 × 3 a3n =1458                    ... iNow, Sn = 728728 = a 3n-13-1 728 = a3n - a21456 = a3n-1 - a  1456= 1458- a                                    From ia = 1458 - 1456 a= 2

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Question 10:

The ratio of the sum of first three terms is to that of first 6 terms of a G.P. is 125 : 152. Find the common ratio.

Answer:

Let a be the first term and r be the common ratio of the G.P.

 S3 =ar3-1r-1  and S6 = ar6-1r-1Then, according to the question       S3S6=ar3-1r-1   a r6-1r-1 125152 = r3-1r6-1125 r6-1 = 152 r3-1125r6-125 = 152r3 - 152125r6 - 152r 3+ 27 = 0Now, let r3 = y  125y2 - 152y + 27 = 0Now, applying the quadatic formulay = -b±b2-4ac2a  y = 152±9604250 y = 152+9604250 or  152-9604250y = 1 or 27125 r3 = 1 or   r3 = 27125But, r = 1 is not possible. r = 271253 = 35

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Question 11:

The 4th and 7th terms of a G.P. are 127 and 1729 respectively. Find the sum of n terms of the G.P.

Answer:

Let a be the first term and r be the common ratio of the G.P.

 a4=127  ar4-1 = 127ar3 = 127  ar32 = 1272a2r6 = 1729  ar6 = 1729a             ... iSimilarly, a7=1729   ar7-1 = 1729ar6 = 1729  ar6= 1729a                        From i             a = 1Putting this in a4=127ar3 = 133r3 =  133  r = 13Now, sum of n terms of the G.P., Sn = arn-1r-1Sn = 11-13n1-13 Sn= 321-13n

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Question 12:

Find the sum : n=110 12n-1+15n+1.

Answer:

n=11012n-1 + 15n+1=n=11012n-1 + n=11015n+1   = 1+12+14+ ... +129 + 152+153+154+ ...+1511=11-12101-12 + 1251-15101-15 =210-129 + 510-14×511

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Question 13:

The fifth term of a G.P. is 81 whereas its second term is 24. Find the series and sum of its first eight terms.

Answer:

Let a be the first term and r be the common ratio of the G.P.

         a2=24 ar2-1=24ar = 24                                                 ...iSimilarly, a5 = 81 ar5-1 = 24ar4 = 8124×r4r= 81                                      From ir3 = 8124   r3 =278r =32Putting r =32 in i3a = 48   a =16So, the geometric series is 16+ 24 + 36+ ...+16328And, S8=16328-132-1 S8=326561-256256=32×6305256=63058

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Question 14:

If S1, S2, S3 be respectively the sums of n, 2n, 3n terms of a G.P., then prove that S12+S22 = S1 (S2 + S3).

Answer:

Let a be the first term and r be the common ratio of the given G.P.

 Sum of n terms, S1=arn-1r-1                 ... 1Sum of 2n terms, S2 =ar2n-1r-1S2 =arn2-12r-1S2 =arn-1rn+1r-1S2 =S1rn+1                                          ....2And, sum of 3n terms, S3= ar3n-1r-1S3=arn3-13r-1S3=arn-1r2n+rn+1r-1S3=S1r2n+rn+1                                ...3

Now, LHS=S12 + S22 =S12 + S1rn+12                   Using 2=S121 + rn+12=S121+r2n+2rn+1=S12r2n+rn+1+rn+1=S1S1r2n+rn+1+S1rn+1=S1S2+S3                              Using 2 and 3=RHSHence proved.

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Question 15:

Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is 1rn.

Answer:

Let a be the first term and r be the common ratio of the G.P.
Sum of the first n terms of the series = a1+a2+a3+ ... +an
Similarly, sum of the terms from n+1th to 2nth term  = an+1 +an+2 + ... + a2n
 Required ratio = a1+a2+a3+ ... +an an+1 +an+2 + ... +a2n  = a +ar+ ... +arn-1arn + arn+1+ ... +ar2n-1=a1-rn1-rarn1-rn1-r = 1rn

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Question 16:

If a and b are the roots of x2 − 3x + p = 0 and c, d are the roots x2 − 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (qp) = 17 : 15.

Answer:

We have,
a +b = 3, ab = p, c + d =12 and cd = q
a, b, c and d form a G.P.
∴ First term = ab = ar, c = ar2 and d = ar3
Then, we have
a + b = 3  and c + d = 12
 a+ar = 3  a( 1+ r ) = 3                            ...iSimilarly, ar2(1+r) = 12              ...iiar21+ra1+r = 123r2 = 4 r = 2 a 1+r = 3  a = 1Now,  p = ab  p =  a × ar = 2And, q = cd q = ar2 × ar3 = 25 = 32q+pq-p=32+232-2=3430=1715



Page No 20.29:

Question 17:

How many terms of the G.P. 3, 32,34..... are needed to give the sum 3069512?

Answer:

Here, a = 3 and Common ratio, r =12 And, Sn = 3069512Sn = 31-12n1-123069512=31-12n1-12 3069512= 6 1-12n30693072 = 1 -12n 12n = 1 - 30693072 12n=330722n = 30723 2n= 1024 2n= 210 n = 10

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Question 18:

A person has 2 parents, 4 grandparents, 8 great grandparents, and so on. Find the number of his ancestors during the ten generations preceding his own.

Answer:

Here, the ancestors of the person form the G.P.  2, 4, 8, 16, ........
Now, first term, a = 2 
And,  r = 2
∴ Number of  his ancestors during the ten generations preceding his own, S10 = 2210-12-1 = 2 1024-1=2046      

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Question 19:

If S1, S2, ..., Sn are the sums of n terms of n G.P.'s whose first term is 1 in each and common ratios are 1, 2, 3, ..., n respectively, then prove that S1 + S2 + 2S3 + 3S4 + ... (n − 1) Sn = 1n+ 2n + 3n + ... + nn.

Answer:

Given: S1, S2, ..., Sn are the sum of n terms of an G.P. whose first term is 1 in each case and the common ratios are 1, 2, 3, ..., n. S1=1+1+1+ ... n terms = n               ... 1   S2=12n-12-1 =2n-1                                  ... 2   S3=13n-13-1 =3n-12                               ... 3   S4=14n-14-1 =4n-13                        ... 4....   Sn=1nn-1n-1 =nn-1n-1                      ............nNow, LHS= S1+S2+2S3+3S4 + ... +n-1Sn=n+2n-1+3n-1+4n-1+ ... +nn-1              Using 1, 2, 3, ..., n=n+2n+3n+4n+ ... +nn-1+1+1+ ... +n-1 times=n+2n+3n+4n+ ... +nn-n-1=n+2n+3n+4n+ ... +nn-n+1=1+2n+3n+4n+ ... +nn=1n+2n+3n+4n+ ... +nn=RHSHence proved.

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Question 20:

A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying the odd places. Find the common ratio of the G.P.

Answer:

Let there be 2n terms in the given G.P. with the first term being a and the common ratio being r.
According to the question
Sum of all the terms = 5 (Sum of the terms occupying the odd places)
a1 + a2 + ... +a2n = 5 a1+a3+a5+ ... +a2n-1a + ar + ... +ar2n-1 =  5 a+ar2 + ... +ar2n-2a1-r2n1-r = 5a1-r2n1-r2  1+r = 5  r = 4

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Question 21:

Let an be the nth term of the G.P. of positive numbers. Let n=1100a2n=α and n=1100a2n-1=β, such that α ≠ β. Prove that the common ratio of the G.P. is α/β.

Answer:

Let a be the first term and r be the common ratio of the G.P.

 n=1100a2n = α  and n=1 100a2n-1 =β a2+a4+ ... +a200 = α  and  a1+a3+ ... +a199 = βar + ar3 + ... +ar199 =α and  a+ar2+ ... +ar198 = βar1-r21001-r2 = α  and   a1-r21001-r2 = βNow, dividing α by βαβ = ar1-r21001-r2  a1-r21001-r2 = arr=r r = αβ

Page No 20.29:

Question 22:

Find the sum of 2n terms of the series whose every even term is 'a' times the term before it and every odd term is 'c' times the term before it, the first term being unity.

Answer:

Let the given series be a1 + a2 + a3 + a4+ ... +a2n.Now, it is given that a1 = 1, a2 = aa1, a3 = ca2, a4 = aa3, a5 =ca4 and so on. a1 = 1a1 = 1, a2 = a, a3 = ac, a4 = a2c, a5=a2c2, a6 = a3c2,..... Sum of the 2n terms of the series, Sn= a1 + a2 + a3 + a4 + ... + a2n= 1 + a + ac +a2c + a2c2 + ... +2n terms=1+a + ac1+a +a2c21+a + ... + n terms=1+a1-acn1-ac = 1+a acn-1ac-1                       



Page No 20.39:

Question 1:

Find the sum of the following series to infinity:

(i) 1-13+132-133+134+...

(ii) 8 + 42 + 4 + ... ∞

(iii) 2/5 + 3/52 +2/53 + 3/54 + ... ∞.

(iv) 10 − 9 + 8.1 − 7.29 + ... ∞

(v) 13+152+133+154+135+156+...

Answer:

i In the given G.P., first term, a=1 and common ratio, r=-13Hence, the sum S to infinity is given by S = a1-r=11--13=34.ii In the given G.P., first term, a=8 and common ratio, r=12Hence, the sum S to infinity is given by S = a1-r=81-12=2+2.iii We have: 25+352+253+354 . = 25+253+ ..+352+354+ ...=An infinite G.P. with a= 25 and r=125+An infinite G.P.with a=325 and r=125= 251-125+ 351-125= 252425+352425=1024+324=1324

Page No 20.39:

Question 2:

Prove that: (91/3 . 91/9 . 91/27 ... ∞) = 3.

Answer:

LHS=913.919.9127 ... =913+19127.=9131-13=9131-13=9=3=RHS



Page No 20.40:

Question 3:

Prove that: (21/4 . 41/8 . 81/16. 161/32 ... ∞) = 2.

Answer:

LHS=214.428.8316.16432 ...=214+28+316316432.=2122+223+324+425 + ...=21221+22+322+423 ...=212211-12+1.121-122=21222+2=21=2=RHS

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Question 4:

If Sp denotes the sum of the series 1 + rp + r2p + ... to ∞ and sp the sum of the series 1 − rp + r2p − ... to ∞, prove that Sp + sp = 2 . S2p.

Answer:

We have:Sp= 1+rp+r2p+ ...  Sp=11-rpSimilarly, sp= 1-rp+r2p- ...  sp=11--rp=11+rpNow, SP+sp = 11-rp+11+rp = 1-rp+1+rp1-r2p21-r2p = 2S2P SP+sp = 2S2P

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Question 5:

Find the sum of the terms of an infinite decreasing G.P. in which all the terms are positive, the first term is 4, and the difference between the third and fifth term is equal to 32/81.

Answer:

Let r be the common ratio of the given G.P. a = 4Sum of the geometric ifinite series:S= 4+4r+4r2+ ... Now, S = 41-r                    .......iThe difference between the third and fifth term is 3281.a3 -a5 = 32814r2-4r4 = 3281 4r2-r4=328181r4-81r2+8 =0                    .......iiNow, let r2 = yLet us put this in ii. 81r4-81r2+8 =081y2-81y+8 = 081y2-72y-9y+8 = 09y9y-1-89y-1 = 09y-89y-1y=19, 89Putting y = r2, we get r=13 and 223Substituting  r=13 and 223 in i:S= 41-13=122=6Similarly, S = 41-223=123-22 S = 6, 123-22

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Question 6:

Express the recurring decimal 0.125125125 ... as a rational number.

Answer:

Let the rational number S be 0.125. S=0.125=0.125+0.000125+0.000000125+0.000000000125+... S=0.1251+10-3+10-6+10-9 + ... Clearly, S is a geometric series with the first term, a, being 1 and the common ratio, r, being 10-3. S=11-rS=0.12511-10-3S=125999

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Question 7:

Find the rational number whose decimal expansion is 0.423.

Answer:

Let the rational number S be 0.423. S=0.423=0.4+0.023+0.00023+0.0000023+ ... S=0.4+0.0231+10-2+10-4+ ... Clearly, S is a geometric series with the first term, a, being 1 and the common ratio, r, being 10-2. S=0.4+0.02311-10-2S=0.4+2.399S=419990

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Question 8:

Find the rational numbers having the following decimal expansions:
(i) 0.3
(ii) 0.231
(iii) 3.52
(iv) 0.68

Answer:

(i) 0.3Let S=0.3S=0.3+0.03+0.003+0.0003+0.00003+ ...  S=0.31+10-1+10-2+10-3+10-4+ ... S is a geometric series with the first term, a, being 1 and the common ratio, r, being 10-1. S=11-rS=0.311-10-1S=39=13

(ii) 0.231Let S = 0.231S=0.231+0.000231+0.000000231+ ...  S=0.2311+10-3+10-6+ ... It is a G.P. S=0.23111-10-3S=231999

(iii) 3.52Let S=3.52S=3.5+0.02+0.002+0002+0.00002+ ... S=3.5+0.021+10-1+10-2+10-3+10-4+ ... It is a G.P.  S=3.5+0.0211-10-1S=3.5+0.29S=31790

(iv) 0.68Let S=0.68S=0.6+0.08+0.008+0.0008+0.00008+ ... S=0.6+0.081+10-1+10-2+10-3+ ... It is a G.P. S=0.6+0.0811-10-1S=0.6+0.89S=6.29S=6290=3145

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Question 9:

One side of an equilateral triangle is 18 cm. The mid-points of its sides are joined to form another triangle whose mid-points, in turn, are joined to form still another triangle. The process is continued indefinitely. Find the sum of the (i) perimeters of all the triangles. (ii) areas of all triangles.

Answer:

According to the midpoint theorem, the sides of each triangle formed by joining the midpoints of an equilateral triangle are half of the sides of the equilateral triangle. In other words, the triangles formed are equilateral triangles with sides 18 cm, 9 cm, 4.5 cm, 2.25 cm, ...

(i) Sum of the perimeters of all the triangles, P= 3×18+3×9+3×4.5+3×2.25+... P=3×18+9+4.5+2.25+... It is a G.P. with a=18 and r=12. P=3×181-12P=3×36=108 cm

(ii) Sum of the areas of all the triangles, A=34182+3492+344.52+ ... A=34182+92+4.52+ ... It is a G.P. with a= 182 and r=14. A=34 1821-14A=33×324A=1083 cm2

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Question 10:

Find an infinite G.P. whose first term is 1 and each term is the sum of all the terms which follow it.

Answer:

Here, first term, a = 1
Common ratio = r

 an = an+1 + an+2+ an+3+.....  nNarn-1 = arn+ arn-1+.....rn-1 = rn1-r                        Putting a =1rn-1 1-r = rn 1-r = r2r = 1    r =12Thus, the infinte G.P is 1, 12, 14, ... .

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Question 11:

The sum of first two terms of an infinite G.P. is 5 and each term is three times the sum of the succeeding terms. Find the G.P.

Answer:

Let the first term be a and the common difference be r.

 a1 + a2 = 5  a + ar = 5                        ...iAlso, an= 3 an+1+an+2+an+3+ ...   n Narn-1 = 3 arn+1+arn+2+arn+3+ ... arn-1 = 3arn1-r  1-r = 3r4r = 1 r =14Putting r =14 in i:a + a4 = 55a = 20  a = 4Thus, the G.P. is 4, 1, 14, 116, ....

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Question 12:

Show that in an infinite G.P. with common ratio r (|r| < 1), each term bears a constant ratio to the sum of all terms that follow it.

Answer:

Let us take a G.P. with terms a1, a2, a3, a4, ... and common ratio rr<1.Also, let us take the sum of all the terms following each term to be S1, S2, S3, S4, ...Now, S1=a21-r=ar1-r,S2=a31-r=ar21-r,S3=a41-r=ar31-r,...a1S1=aar1-r=1-rr,a2S2=arar21-r=1-rr,a3S3=ar2ar31-r=1-rr,...It is clearly seen that the ratio of each term to the sum of all the terms following it is constant.

Page No 20.40:

Question 13:

If S denotes the sum of an infinite G.P. S1 denotes the sum of the squares of its terms, then prove that the first term and common ratio are respectively 2SS1S2+S1 and S2-S1S2+S1.

Answer:

 S=a1-r                .......(i)And, S1=a21-r2           S1=a21-r1+r    .......(ii)Now, putting the value of a in equation (ii) from equation (i):S1=S21-r21-r1+rS1=S21-r1+rS11+r=S21-rrS1+S2=S2-S1r=S2-S1S1+S2Putting the value of r in equation (i):a=S1-ra=S1-S2-S1S1+S2a=SS1+S2-S2-S1S1+S2a=2SS1S1+S2



Page No 20.45:

Question 1:

If a, b, c are in G.P., prove that log a, log b, log c are in A.P.

Answer:

a ,b and c are in G.P.
 b2 = ac 
Now, taking log on both the sides:logb2 = log ac 2log b = log a + log cThus, log a, log b and log c are in A.P.

Page No 20.45:

Question 2:

If a, b, c are in G.P., prove that
1loga m, 1logb m,1logc m are in A.P.

Answer:

a, b, c are in G.P.
 b2 = ac Now taking logm on both the sides:logmb2 = logmac2logmb = logma  + logmc2logbm = 1logam +1logcmThus, 1logam, 1logbm and 1logcm are in A.P.  

Page No 20.45:

Question 3:

Find k such that k + 9, k − 6 and 4 form three consecutive terms of a G.P.

Answer:

k, k + 9, k−6 are in G.P.
 k-62=4k+9k2+36-12k = 4k + 36k2 - 16k = 0k k-16 = 0k=0, 16
But, k = 0 is not possible.
k = 16

Page No 20.45:

Question 4:

Three numbers are in A.P. and their sum is 15. If 1, 3, 9 be added to them respectively, they form a G.P. Find the numbers.

Answer:

Let the first term of an A.P. be a and its common difference be d.

a1+a2+a3=15a+a+d+a+2d=153a + 3d = 15  a+d = 5                     .......(i)Now, according to the question:a + 1, a+d+3 and a+2d+9 are in G.P.a+d+32 = a+1a+2d+95-d+d+32 = 5-d+1 5-d+2d+9          From (i)    82 = 6-d14+d64  = 84 + 6d-14d -d2d2+8d-20=0d-2d+10 = 0d=2, -10Now, putting d = 2, -10 in equation (i), we get, a = 3, 15, respectively.Thus, for a = 3 and d=2, the A.P. is 3, 5, 7.And, for a = 15 and d=-10, the A.P. is 15 , 5, -5.    

Page No 20.45:

Question 5:

The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.

Answer:

Let the first term of an A.P is a and its common difference be d.

 a1+a2+a3=21a+a+d+a+2d=213a + 3d = 21  a+d = 7                     ...(i)Now, according to the question:a , a+d-1 and a+2d+1 are in G.P.a+d-12 = aa+2d+17+a-a-12 = a a+27-a+1      62 = a15-a36  = 15a -a2a2-15a+36=0a-3a-12 = 0a=3, 12Now, putting a = 2, 12 in equation (i), we get d = 5, -5, respectively.Thus, for a = 2 and d=5, the numbers are  2, 7 and 12.And, for a = 12 and d=-5, the numbers are 12 ,7 and 2.    

Page No 20.45:

Question 6:

The sum of three numbers a, b, c in A.P. is 18. If a and b are each increased by 4 and c is increased by 36, the new numbers form a G.P. Find a, b, c.

Answer:

Let the first term of the A.P. be a and the common difference be d.
a = a , b = a + d and c = a + 2d
a+b+c=18a+a+d+a+2d=183a + 3d = 18  a+d = 6                     .......(i)Now, according to the question, a + 4, a+d+4 and a+2d+36 are in G.P. a+d+42 = a+4a+2d+366-d+d+42 =6-d+4 6-d+2d+36       102 = 10-d42+d100  = 420 + 10d-42d -d2d2+32d-320=0d+40d-8 = 0d=8, -40Now, putting d = 8, -40 in equation (i), we get, a = -2, 46, respectively.For a = -2 and d=8, we have: a = -2 , b = 6 , c = 14And, for a = 46 and d=-40, we have: a = 46 , b = 6 , c = -34    



Page No 20.46:

Question 7:

The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P. Find the numbers.

Answer:

Let the first term of a G.P be a and its common ratio be r.
  a1+a2+a3=56a+ar+ar2 = 56 a 1+r+r2 = 56                                         a=561+r+r2                                                                  .......i                Now, according to the question:a-1, ar-7 and ar2-21 are in A.P. 2ar-7 = a-1 + ar2 - 212ar - 14 = ar2+a-22ar2-2ar+a-8  = 0a1-r2 = 8a = 81-r2                                                                  .......iiEquating (i) and (ii):81-r2=561+r+r281+r+r2 = 561+r2-2r 1+r+r2 =7 1+r2-2r1+r+r2 =7+7r2-14r6r2-15r+6=0    32r2-5r+2 = 02r2-4r-r+2=02r(r-2)-1(r-2)=0(r-2)(2r-1)=0r=2, 12When r=2, a=8.        [Using (ii)]And, the required numbers are 8, 16 and 32.When r=12, a=32.   [Using (ii)]And, the required numbers are 32, 16 and 8.

Page No 20.46:

Question 8:

If a, b, c are in G.P., prove that:

(i) a (b2 + c2) = c (a2 + b2)

(ii) a2b2c21a3+1b3+1c3=a3+b3+c3

(iii) (a+b+c)2a2+b2+c2=a+b+ca-b+c

(iv) 1a2-b2+1b2=1b2-c2

(v) (a + 2b + 2c) (a − 2b + 2c) = a2 + 4c2.

Answer:

a, b and c are in G.P.
 b2=ac           .......(1)

(i) LHS=ab2+c2=ab2+ac2=aac+cb2         Using (1)=ca2+b2=RHS

(ii) LHS=a2b2c21a3+1b3+1c3=b2c2a+a2c2b+a2b2c=acc2a+b22b+a2acc        Using (1)=a3+b3+c3=RHS

(iii) LHS=a+b+c2a2+b2+c2=a+b+c2a2-b2+c2+2b2=a+b+c2a2-b2+c2+2ac           Using (1)=a+b+c2a+b+ca-b+c         a+b+ca-b+c=a2-b2+c2+2ac=a+b+ca-b+c =RHS 

(iv) LHS=1a2-b2+1b2=b2+a2-b2a2-b2b2=a2a2b2-b4=a2a2ac-ac2=1ac-c2=1b2-c2=RHS

(v) LHS =a+2b+2ca-2b+2c=a2-4b2+4c2+4ac=a2-4ac+4c2+4ac                Using (1)=a2+4c2=RHS

Page No 20.46:

Question 9:

If a, b, c, d are in G.P., prove that:

(i) ab-cdb2-c2=a+cb

(ii) (a + b + c + d)2 = (a + b)2 + 2 (b + c)2 + (c + d)2

(iii) (b + c) (b + d) = (c + a) (c + d)

Answer:

a, b, c and d are in G.P.

 b2=acbc=adc2=bd              .......(1)

(i) LHS=ab-cdb2-c2=ab-cdac-bd           Using (1)=ab-cdbac-bdb  =ab2-bcdac-bdb=aac-cc2ac-bdb       Using (1)=a2c-c3ac-bdb=ca2-c2ac-bdb=a+cac-c2ac-bdb=a+cac-bdac-bdb          Using (1)=a+cb=RHS

(ii) LHS=a+b+c+d2=a+b2+2a+bc+d+c+d2=a+b2+2ac+ad+bc+bd+c+d2=a+b2+2b2+bc+bc+c2+c+d2         Using (1)=a+b2+2b+c2+c+d2=RHS

(iii) LHS=b+cb+d=b2+bd+bc+cd=ac+c2+ad+cd           Using (1)=ca+c+da+c=c+ac+d = RHS

Page No 20.46:

Question 10:

If a, b, c are in G.P., prove that the following are also in G.P.:
(i) a2, b2, c2
(ii) a3, b3, c3
(iii) a2 + b2, ab + bc, b2 + c2

Answer:

a, b and c are in G.P.
b2=ac        .......(1)
(i) b22=ac2         Using (1)b22=a2c2Therefore, a2, b2 and c2 are also in G.P.

(ii) b32=b23=ac3         Using (1)b32=a3c3Therefore, a3, b3 and c3 are also in G.P.

(iii) ab+bc2=ab2+2ab2c+bc2ab+bc2=ab2+ab2c+ab2c+bc2ab+bc2=a2b2+acac+b2b2+b2c2           Using (1)ab+bc2=a2b2+c2+b2b2+c2ab+bc2=b2+c2a2+b2Therefore, a2+b2, b2+c2 and ab+bc are also in G.P.

Page No 20.46:

Question 11:

If a, b, c, d are in G.P., prove that:

(i) (a2 + b2), (b2 + c2), (c2 + d2) are in G.P.

(ii) (a2b2), (b2c2), (c2d2) are in G.P.

(iii) 1a2+b2,1b2-c2,1c2+d2 are in G.P.

(iv) (a2 + b2 + c2), (ab + bc + cd), (b2 + c2 + d2) are in G.P.

Answer:

a, b, c and d are in G.P.

 b2=acad=bc       c2=bd       .......(1)

(i) b2+c22=b22+2b2c2+c22b2+c22=ac2+b2c2+b2c2+bd2           Using (1)b2+c22=a2c2+a2d2+b2c2+b2d2            Using (1)b2+c22=a2c2+d2+b2c2+d2 b2+c22=a2+b2c2+d2Therefore, a2+b2, c2+d2 and b2+c2 are also in G.P.

(ii) b2-c22=b22-2b2c2+c22b2-c22=ac2-b2c2-b2c2+bd2            Using (1)b2-c22=a2c2-b2c2-a2d2+b2d2             Using (1)b2-c22=c2a2-b2-d2a2-b2b2-c22=a2-b2c2-d2Therefore, a2-b2, b2-c2 and c2-d2 are also in G.P.

(iii)  1b2+c22=1b22+2b2c2+1c221b2+c22=1ac2+1b2c2+1b2c2+1bd2           Using (1)1b2+c22=1a2c2+1a2d2+1b2c2+1b2d2            Using (1)1b2+c22=1a21c2+1d2+1b21c2+1d21b2+c22=1a2+b21c2+1d2Therefore, 1b2+c2, 1c2+d2 and 1b2+c2 are also in G.P.

(iv) ab+bc+cd2=ab2+bc2+cd2+2ab2c+2bc2d+2abcdab+bc+cd2=a2b2+b2c2+c2d2+ab2c+ab2c+bc2d+bc2d+abcd+abcdab+bc+cd2=a2b2+b2c2+c2d2+b2b2+acac+c2c2+bdbd+bcbc+adad        Using (1)ab+bc+cd2=a2b2+a2c2+a2d2+b4+b2c2+b2d2+c2b2+c4+c2d2ab+bc+cd2=a2b2+c2+d2+b2b2+c2+d2+c2b2+c2+d2ab+bc+cd2=b2+c2+d2a2+b2+c2Therefore, a2+b2+c2, ab+bc+cd and b2+c2+d2 are also in G.P.

Page No 20.46:

Question 12:

If (a b), (bc), (ca) are in G.P., then prove that (a + b + c)2 = 3 (ab + bc + ca)

Answer:

 a-b, b-c and  c-a are in G.P. b-c2=a-bc-ab2-2bc+c2=ac-bc+ab-a2a2+b2+c2=ab+bc+ca            .......(i)Now, LHS=a+b+c2=a2+b2+c2+2ab+2bc+2ca=ab+bc+ca+2ab+2bc+2ca              Using (i)=3ab+3bc+3ca=3ab+bc+ca=RHS

Page No 20.46:

Question 13:

If a, b, c are in G.P., then prove that: a2+ab+b2bc+ca+ab=b+ac+b

Answer:

a, b and c are in G.P. b2=ac                     ........(i)Now, LHS=a2+ab+b2bc+ca+ab=a2+ab+acbc+b2+ab                   Using (i)=aa+b+cbc+b+a=ab=1rHere, r = common ratioRHS=b+ac+b=ar+aar2+ar=a(r+1)ar(r+1)=1r LHS = RHS

Page No 20.46:

Question 14:

If the 4th, 10th and 16th terms of a G.P. are x, y and z respectively. Prove that x, y, z are in G.P.

Answer:

 a4=xar3=xAlso, a10=yar9=yAnd, a16=zar15=z  yx=ar9ar3=r6and zy=ar15ar9=r6yx=zyTherefore, x, y and z are in G.P.

Page No 20.46:

Question 15:

If a, b, c are in A.P. and a, b, d are in G.P., then prove that a, ab, dc are in G.P.

Answer:

a, b and c are in A.P. 2b=a+c        .......(i)Also, a, b and d are in G.P. b2=ad           .......(ii)Now, a-b2=a2-2ab+b2=a2-aa+c+ad            Using (i) and (ii)=ad-ac=ad-ca-b2=ad-cTherefore, a, a-b and d-c are in G.P.

Page No 20.46:

Question 16:

If pth, qth, rth and sth terms of an A.P. be in G.P., then prove that pq, qr, rs are in G.P.

Answer:

Here, ap=a + p-1daq= a + q-1dar = a + r-1d as= a + s-1dIt is given that ap, aq, ar and as are in G.P. aqap=araq=aq-arap-aq=q-rp-q                           .......(i)Similarly, araq=asar=ar-asaq-ar=r-sq-r                        .......(ii)Using i and ii:q-rp-q=r-sq-r,  Therefore, p-q, q-r and r-s are in G.P.

Page No 20.46:

Question 17:

If 1a+b,12b,1b+c are three consecutive terms of an A.P., prove that a, b, c are the three consecutive terms of a G.P.

Answer:

Here, 1a+b, 12b and 1b+c are in A.P.

 2×12b=1a+b+1b+c1b=b+c+a+ba+bb+ca+bb+c =b2b+a+cab+ac+b2+bc = 2b2+ab+bc2b2 - b2 = acb2 = acThus, a, b and c are in G.P.

Page No 20.46:

Question 18:

If xa = xb/2zb/2 = zc, then prove that 1a,1b,1c are in A.P.

Answer:

Here, xa= xzb2 = zcNow, taking log on both the sides:logxa=logxzb2=logzcalog x =b2 logxz = c log zalog x =b2log x + b2log z = c log zalog x =b2log x + b2log z andb2log x +b2log z= c log za-b2log x=b2 log z andb2log x=c-b2log zlog x log z=b2a-b2 and log x log z=c-b2b2 b2a-b2=c-b2b2b24=ac-ab2-bc2+b242ac=ab+bc2b = 1a+1cThus, 1a, 1b and 1c are in A.P.

Page No 20.46:

Question 19:

If a, b, c are in A.P., b,c,d are in G.P. and 1c,1d,1e are in A.P., prove that a, c,e are in G.P.

Answer:

a, b and c are in A.P. 2b=a+c               .......(i)Also, b, c and d are in G.P. c2=bd                      .......(ii)And 1c, 1d and 1e are in A.P. 2d=1c+1e d=2cec+e                     .......(iii) c2=bd              From (ii)   c2=a+c  22cec+e          Using (i) and (iii)c2c+e=cea+cc2+ce=ae+ecc2=aeTherefore, a, c and e are also in G.P.

Page No 20.46:

Question 20:

If a, b, c are in A.P. and a, x, b and b, y, c are in G.P., show that x2, b2, y2 are in A.P.

Answer:

a, b and c are in A.P. 2b=a+c               .......(i)a, x and b are in G.P. x2=ab                   .......(ii)And, b, y and c are also in G.P. y2=bc                    .......(iii)Now, putting the values of a and c:2b=x2b+y2b2b2=x2+y2Therefore, x2, b2 and y2 are also in A.P.

Page No 20.46:

Question 21:

If a, b, c are in A.P. and a, b, d are in G.P., show that a, (ab), (dc) are in G.P.

Answer:

a, b and c are in A.P. 2b=a+c               .......(i)Also, a, b and d are in G.P. b2=ad                   .......(ii)Now, a-b2=a2-2ab+b2 a-b2=a2-aa+c+ad           Using (i) and (ii) a-b2=a2-a2-ac+ad a-b2=ad-ac a-b2=a(d-c)Therefore, a, a-b and (d-c) are in G.P.

Page No 20.46:

Question 22:

If a, b, c are three distinct real numbers in G.P. and a + b + c = xb, then prove that either x < −1 or x > 3.

Answer:

Let r be the common ratio of the given G.P. b=ar and c=ar2Now, a+b+c=bxa+ar+ar2=arxr2+1-xr+1=0 r is always a real number. D01-x2-40x2-2x-30x-3x+10x>3 or x<-1 and x3 or -1     a, b and c are distinct real numbers

Page No 20.46:

Question 23:

If pth, qth and rth terms of an A.P. and G.P. are both a, b and c respectively, show that ab-cbc-aca-b=1.

Answer:

Let A be the first term and D be the common difference of the AP. Therefore,

ap=A+p-1D=a                  .....1aq=A+q-1D=b                  .....2ar=A+r-1D=c                    .....3

Also, suppose A' be the first term and R be the common ratio of the GP. Therefore,

ap=A'Rp-1=a                  .....4aq=A'Rq-1=b                  .....5ar=A'Rr-1=c                    .....6

Now,

Subtracting (2) from (1), we get

A+p-1D-A-q-1D=a-bp-qD=a-b                                   .....7

Subtracting (3) from (2), we get

A+q-1D-A-r-1D=b-cq-rD=b-c                                   .....8

Subtracting (1) from (3), we get

A+r-1D-A-p-1D=c-ar-pD=c-a                                   .....9

ab-cbc-aca-b

=A'Rp-1q-rD×A'Rq-1r-pD×A'Rr-1p-qD                 [Using (4), (5), (6), (7), (8) and (9)]

=A'q-rDRp-1q-rD×A'r-pDRq-1r-pD×A'p-qDRr-1p-qD

=A'q-rD+r-pD+p-qD×Rp-1q-rD+q-1r-pD+r-1p-qD

=A'q-r+r-p+p-qD×Rpq-pr-q+r+qr-pq-r+p+pr-qr-p+qD=A'0×R0=1×1=1



Page No 20.54:

Question 1:

Insert 6 geometric means between 27 and 181.

Answer:

Let the 6 G.M.s between 27 and 181 be G1, G2, G3, G4, G5 and G6.Thus, 27, G1, G2, G3, G4, G5, G6 and 181 are in G.P. a=27, n=8 and a8=181 a8=181ar7=181r7=181×27r7=137r=13  G1=a2=ar=2713=9G2=a3=ar2=27132=3G3=a4=ar3=27133=1G4=a5=ar4=27134=13G5=a6=ar5=27135=19 G6=a7=ar6=27136=127

Page No 20.54:

Question 2:

Insert 5 geometric means between 16 and 14.

Answer:

Let the 5 G.M.s betweem 16 and 14 be G1, G2, G3, G4 and G5.16, G1, G2, G3, G4, G5, 14a=16, n=7 and a7=14 a7=14ar6=14r6=14×16r6=126r=12 G1=a2=ar=1612=8G2=a3=ar2=16122=4G3=a4=ar3=16123=2G4=a5=ar4=16124=1G5=a6=ar5=16125=12

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Question 3:

Insert 5 geometric means between 329and812.

Answer:

Let the 5 G.M.s between 329 and 812 be G1, G2, G3, G4 and G5.329, G1, G2, G3, G4, G5, 812a=329, n=7 and a7=812 a7=812ar6=812r6=812×932r6=326r=32  G1=a2=ar=32932=163G2=a3=ar2=329322=8G3=a4=ar3=329323=12G4=a5=ar4=329324=18G5=a6=ar5=329325=27



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Question 4:

Find the geometric means of the following pairs of numbers:
(i) 2 and 8
(ii) a3b and ab3
(iii) −8 and −2

Answer:

(i) Let the G.M. between 2 and 8 be G.Then, 2, G and 8 are in G.P. G2=2×8G2=16G=±16G=±4

(ii) Let the G.M. between a3b and ab3 be G.Then, a3b, G and ab3 are in G.P. G2=a3b×ab3G2=a4b4G=a4b4G=a2b2

(iii) Let the G.M. between -8 and -2 be G.Then, -8, G and -2 are in G.P. G2=-8-2G2=16G=±16G=4, -4

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Question 5:

If a is the G.M. of 2 and 14, find a.

Answer:

a is the G.M. between 2 and 14. a2=2×14a2=12a=12a=12

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Question 6:

Find the two numbers whose A.M. is 25 and GM is 20.

Answer:

Let A.M. and G.M. between the two numbers a and b be A and G, respectively. A=25a+b2=25a+b=50              .......(i)Also G= 20ab=20ab=400                 .......(ii)Now, putting the value of a in (ii):(50-b)b=400b2-50b+400=0b2-10b-40b+400=0bb-10-40b-10=0b-10b-40=0b=10, 40If b=10, then, a=400.And, if b=40, then a=10.Thus, the two numbers are 10 and 40.

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Question 7:

Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Answer:

Let the roots of the quadratic equation be a and b. A=a+b2 a+b=2A        ........(i)Also, G2=ab        .......(ii)The quadratic equation having roots a and b is given by x2-(a+b)x+ab=0. x2-2Ax+G2=0              Using (i) and (ii)

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Question 8:

The sum of two numbers is 6 times their geometric means, show that the numbers are in the ratio (3+22) : (3-22).

Answer:

Let the two numbers be a and b.Let the geometric mean between them be G.We have:a+b=6GBut, G=ab a+b=6aba+b2=6ab2a2+2ab+b2=36aba2-34ab+b2=0Using the quadratic formula:a=--34b±-34b2-4×1×b22×1a=34b±b1156-42a=b34±11522ab=34±2422ab=17+122           a and b are positive numbersab=3+8+2×3×22ab=3+222ab=3+2223-223-22ab=3+229-83-22ab=3+223-22a:b=3+22:3-22

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Question 9:

If AM and GM of roots of a quadratic equation are 8 and 5 respectively, then obtain the quadratic equation.

Answer:

Let the roots of the quadratic equation be a and b. AM=8 a+b2=8a+b=16             ........(i)Also, G=5ab=5ab=52ab=25                ........(ii)Now, the quadratic equation is given by x2-(a+b)x+ab=0x2-16x+25=0

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Question 10:

If AM and GM of two positive numbers a and b are 10 and 8 respectively, find the numbers.

Answer:

AM=10 a+b2=10a+b=20               ........(i)Also, G=8 ab=8ab=82ab=64                   ........(ii)Using (i) and (ii): a20-a=64a2-20a+64=0a2-16a-4a+64=0aa-16-4a-16=0a-16a-4=0a=4, 16If a=4, then b=16.And, if a=16, then b=4.

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Question 11:

Prove that the product of n geometric means between two quantities is equal to the nth power of a geometric mean of those two quantities.

Answer:

Let G1, G2, G3, G4, ..., Gn be n G.M.s between a and b.Then, a, G1, G2, G3, G4, ..., Gn, b is a G.P.Let r be the common ratio. b=an+2=arn+1r=ba1n+1 G1=a2=arG2=a3=ar2G3=a4=ar3...Gn=an+1=arnAlso, let G be the G.M. between a and b. G2=abNow, G1×G2×G3×G4× ... × Gn = ar×ar2×ar3×ar4× ... ×arn=an×r1+2+3+4+......+n=an×rnn+12=an×ba1n+1nn+12=an×ban2=an2×bn2=abn2=abn=Gn G1×G2×G3×G4× ... × Gn=Gn

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Question 12:

If the A.M. of two positive numbers a and b (a > b) is twice their geometric mean. Prove that: a : b = (2+3) : (2-3).

Answer:

 AM=2GM a+b2=2aba+b=4abSquaring both the sides:a+b2=4ab2a2+2ab+b2=16aba2-14ab+b2=0Using the quadratic formula:a=--14b+-14b2-4×1×b22×1     a is positive numbera=14b+2b49-12a=b7+43ab=7+43ab=4+3+2×2×3ab=2+32ab=2+322-32-3ab=2+34-32-3 ab=2+32-3

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Question 13:

If one A.M., A and two geometric means G1 and G2 inserted between any two positive numbers, show that G12G2+G22G1=2A.

Answer:

Let the two positive numbers be a and b.a, A and b are in A.P. 2A=a+b                 ........(i)Also, a, G1, G2 and b are in G.P. r=ba13Also, G1=ar and G2=ar2.          ........(ii)Now, LHS=G12G2+G22G1=ar2ar2+ar22ar        Using (ii)=a+ar3=a+aba133=a+aba=a+b=2A=RHS                         Using (i)



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Question 1:

If the fifth term of a G.P. is 2, then write the product of its 9 terms.

Answer:

Here, a5 = 2 
ar4 = 2
Product of the nine terms, i.e. a, ar, ar2, ar3, ar4, ar5, ar6, ar7 and ar8:
a×ar8ar×ar7ar2×ar6ar3×ar5ar4 = ar49ar4 = 2Required product = 29 = 512                                                       

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Question 2:

If (p + q)th and (pq)th terms of a G.P. are m and n respectively, then write is pth term.

Answer:

Here, p + qth term = m   arp + q-1 = m           .......iAnd, p - qth  term = n     arp - q-1 = n            .......iiDividing i by ii:arp + q-1arp - q-1=mn    r2q = mnrq = mnNow, from i:arp-1×rq = marp-1 × mn  = marp-1 = m ×nmarp-1 = mnmThus, the pth term is mnm.

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Question 3:

If logxa, ax/2 and logbx are in G.P., then write the value of x.

Answer:

 logx a, ax2 and logb x are in G.P. ax22 = logx a × logb x ax = logba logbx×  logbx ax =  logba Now, by taking loga on both the sides: x = logalogba

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Question 4:

If the sum of an infinite decreasing G.P. is 3 and the sum of the squares of its term is 92, then write its first term and common difference.

Answer:

Let us take a G.P. whose first term is and common difference is r.
 S=a1-r       a1-r= 3                                     .......iAnd, sum of the terms of the G.P. a2, ar2, ar22, ... :S'= a21-r2   a21-r2 =92                             .......ii2a2 = 9 1-r2    231-r2 =  9 - 9r2               From i181+r2-2r = 9-9r218-9 + 18r2+9r2 -36r = 027r2-36r+9 = 039r2-12r+3= 0 9r2-12r+3= 09r2 -9r-3r+3= 09rr-1-3r-1=09r-3r-1=0r=13 and r = 1.But, r = 1 is not possible. r = 13Now, putting r =13 in a1-r = 3:a = 31-13a = 3×23 = 2

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Question 5:

If pth, qth and rth terms of a G.P. are x, y, z respectively, then write the value of xqryrpzpq.

Answer:

Let us take a G.P. whose first term is A and common ratio is R.
According to the question, we have:ARp-1 = xARq-1 = yARr-1 = z xq-r yr-p zp-q= Aq-r× Rp-1q-r× Ar-p × Rq-1r-p × Ap-q × Rr-1p-q= Aq-r+r-p+p-q × Rpr-pr-q+r+rq-r+p-pq+pr-p-qr+q = A0 × R0=1 xq-r yr-p zp-q = 1

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Question 6:

If A1, A2 be two AM's and G1, G2 be two GM's between a and b, then find the value of A1+A2G1 G2.

Answer:

It is given that A1 and  A2 are the A.M.s between a and b.Thus, a , A1, A2 and b are in A.P. with common difference d.Here, d = b-a3 A1 = a + b-a3 = 2a+b3and A2  = a + 2b-a3=a+2b3It is also given that G1 and G2 are the G.M.s between a and b.Thus, a , G1, G2 and b are in G.P. with common ratio r.Here, r = ba13  G1 = aba13 = b13 a13 and G2 = aba13 2= b13 a13 A1 + A2G1 G2 = 2a+b3+a+2b3 b13 a13 × b13 a13  = a+bab

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Question 7:

If second, third and sixth terms of an A.P. are consecutive terms of a G.P., write the common ratio of the G.P.

Answer:

Here, second term, a2 = a + dThird term, a3  = a + 2dSixth term, a6  = a + 5d As, a2, a3 and a6 are in G.P. First term of G.P. = a2 = A = a + dSecond term of G.P. = Ar = a + 2dThird term of G.P. =Ar2= a + 5d  a + 2d2= a + d× a + 5d a2+4ad+4d2=a2+6ad+5d22ad+d2=0d(2a+d)=0d=0 or 2a+d=0But, d=0 is not possible. d=-2ar= a + 2da + dr=a+2(-2a)a+(-2a)r=31=3 

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Question 8:

Write the quadratic equation the arithmetic and geometric means of whose roots are A and G respectively.

Answer:

Let the roots of the required quadratic equation be a and b. A = a+b2 and G =abThe equation having a and b as its roots isx2-xa+b+ab = 0x2 -2Ax + G2 = 0                            A =a+b2 and G =ab

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Question 9:

Write the product of n geometric means between two numbers a and b.

Answer:

Let G1, G2, ..., Gn be n geometric means between two quantities a and b.Thus, a, G1, G2, ..., Gn, b is a G.P. Let r be the common ratio of this G.P. r = ba1n+1And,  G1 = ar, G2 = ar2, G3 = ar3, ..., Gn= arnNow, product of n geometric means = G1·G2·G3· ... ·Gn = arar2ar3 ... arn=arar2ar3......arn =anr1+2+3 + ...+ n=an rnn+12 =anba1n+1nn+12 =anban2=an2bn2 =abn2                                                                               

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Question 10:

If a = 1 + b + b2 + b3 + ... to ∞, then write b in terms of a.

Answer:

Here, a=1, b, b2, b3, ...  form an infinite G.P.   S=a =1+b+b2+b3+... = 11-ba = 11-b1-b = 1a b = 1 - 1a b = a-1a



Page No 20.57:

Question 1:

If in an infinite G.P., first term is equal to 10 times the sum of all successive terms, then its common ratio is
(a) 1/10
(b) 1/11
(c) 1/9.
(d) 1/20

Answer:

(b) 111

Let the first term of the G.P. be a.
Let its common ratio be r.
​According to the question, we have:
First term = 10        [Sum of all successive terms]
a = 10ar1-ra - ar = 10ar11ar = ar = a11a=111

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Question 2:

If the first term of a G.P. a1, a2, a3, ... is unity such that 4 a2 + 5 a3 is least, then the common ratio of G.P. is
(a) −2/5
(b) −3/5
(c) 2/5
(d) none of these

Answer:

(a) −25

If the first term is 1, then, the G.P. will be 1, r, r2, r3, ...
 
Now, 5r2+4r=5r2+45r=5r2+45r+425-425=5r+252-45This will be the least when r+25=0, i.e. r=-25.        

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Question 3:

If a, b, c are in A.P. and x, y, z are in G.P., then the value of xbcycazab is
(a) 0
(b) 1
(c) xyz
(d) xa yb zc

Answer:

(b) 1

a, b and c are in A.P. 2b = a + c                                    ........iAnd, x, y and z are in G.P.   y2 = xzNow, xb-c yc-a za-b = xb + a - 2b y2b-a-a za-b                From i=xa-b y2b-aza-b=xza-b xzb-a                                  From ii, y2 = xz=xz0 = 1

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Question 4:

The first three of four given numbers are in G.P. and their last three are in A.P. with common difference 6. If first and fourth numbers are equal, then the first number is
(a) 2
(b) 4
(c) 6
(d) 8

Answer:

(d) 8

The first and the last numbers are equal.Let the four given numbers be p, q, r and p.The first three of four given numbers are in G.P. q2 = p·r                                             ........iAnd, the last three numbers are in A.P. with common difference 6.We have:  First term =qSecond term=r= q+6Third term=p= q + 12Also, 2r = q + pNow, putting the values of p and r in i:q2 = q+12q+6 q2 = q2 +18q +72 18q +72 = 0 q + 4 = 0q = -4Now, putting the value of q in  p = q + 12:p = -4 + 12 = 8

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Question 5:

If a, b, c are in G.P. and a1/x = b1/y = c1/z, then xyz are in
(a) AP
(b) GP
(c) HP
(d) none of these

Answer:

(a) AP

a, b and c are in G.P.  b2 = acTaking log on both the sides:2log b = log a + log c                      ........iNow, a1x = b1y =c1zTaking log on both the sides:log ax = log by = log cz                ........iiNow, comparing i and ii:log ax=log a + log c  2y=log czlog ax=log a + log c  2y and log ax=log czlog a 2y-x=xlog c and log alog c=xzlog alog c=x2y-x and log alog c=xzx2y-x=xz 2y = x + zThus,  x, y and z are in A.P.

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Question 6:

If S be the sum, P the product and R be the sum of the reciprocals of n terms of a GP, then P2 is equal to
(a) S/R
(b) R/S
(c) (R/S)n
(d) (S/R)n

Answer:

(d) SRn

​Sum of n terms of the G.P., S = arn-1r-1Product of n terms of the G.P., P = anrnn-12Sum of the reciprocals of n terms of the G.P., R = 1rn-1a1r -1=rn-1arn-1r-1 P2 =  a2r2n-12n  P2 = arn-1r-1rn-1arn-1r-1n P2 = SRn


Let the first term of the G.P. be a and the common ratio be r.Sum of n terms, S =arn-1r-1Product of the G.P., P=anrnn+12Sum of the reciprocals of n terms, R = 1rn  - 1 a1r - 1 = 1-rnrna1-rr               p2  = a2rn+12n                             p2  = arn-1r-11-rnrna1-rrn =SRn

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Question 7:

The fractional value of 2.357 is
(a) 2355/1001
(b) 2379/997
(c) 2355/999
(d) none of these

Answer:

(c)2355999

2.357¯ = 2.0 + 0.357 + 0.000357 + 0.000000357 + ... 2.357¯  = 2 + 357103+357106+357109+ ... 2.357¯  = 2 + 3571031-11032.357¯ =2 +3579992.357¯ = 2355999

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Question 8:

If pth, qth and rth terms of an A.P. are in G.P., then the common ratio of this G.P. is

(a) p-qq-r

(b) q-rp-q

(c) pqr

(d) none of these

Answer:

(b) q-rp-q

Let a be the first term and d be the common difference of the given A.P.
Then, we have:
pth term, ap= a + p-1dqth term, aq= a + q-1drth term, ar = a + r-1dNow, according to the question the pth, the qth and the rth terms are in G.P. a + q-1d2= a + p-1d×a + r-1da2+2a q-1d+ q-1d2=a2+adr-1+p-1+p-1 r-1d2ad2q-2-r-p+2+d2q2-2q+1-pr+p+r-1=0a2q-r-p+dq2-2q-pr+p+r=0               d cannot be 0a=-q2-2q-pr+p+rd2q-r-p Common ratio, r =aq ap= a + q-1da + p-1d=q2-2q-pr+p+rdp+r-2q+ q-1dq2-2q-pr+p+rdp+r-2q+p-1d=q2-2q-pr+p+r+pq+rq-2q2-p-r+2qq2-2q-pr+p+r+p2+pr-2pq-p-r+2q=pq-pr-q2+qrp2+q2-2pq=pq-r-qq-rp-q2=p-qq-rp-q2=q-rp-q                    

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Question 9:

The value of 91/3 . 91/9 . 91/27 ... upto inf, is
(a) 1
(b) 3
(c) 9
(d) none of these

Answer:

(b) 3913×919×9127× ... =913+19+127+ ... Here, it is a G.P. with a=13 and r=13. 9131-13=912=3

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Question 10:

The sum of an infinite G.P. is 4 and the sum of the cubes of its terms is 92. The common ratio of the original G.P. is
(a) 1/2
(b) 2/3
(c) 1/3
(d) −1/2

Answer:

(a) 1/2

Let the G.P. be a, ar, ar2, ar3, ..., . S=4a1-r=4                         (i)Also, sum of the cubes, S1=92a31-r3=92                  (ii)Putting the value of a from (i) to (ii):4(1-r)31-r3=9264(1-r)31-r3=921-r31-r1+r+r2=92641-r21+r+r2=2316161-2r+r2=231+r+r27r2+55r+7=0Using the quadratic formula:r=-55+552-4×7×72×7r=-55+552-14214r=-55+282914

Disclaimer: None of the given options are correct. This solution has been created according to the question given in the book.

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Question 11:

If the sum of first two terms of an infinite GP is 1 every term is twice the sum of all the successive terms, then its first term is
(a) 1/3
(b) 2/3
(c) 1/4
(d) 3/4

Answer:

(d) 3/4

Let the terms of the G.P. be a, a2, a3, a4, a5, ..., .And, let the common ratio be r.Now, a+a2=1 a+ar=1                        ........(i)Also, a=2a2+a3+a4+a5+...a=2ar+ar2+ar3+ar4+...a=2ar1-r1-r=2r3r=1r=13Putting the value of r in (i):a+a3=14a3=14a=3a=34

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Question 12:

The nth term of a G.P. is 128 and the sum of its n terms  is 225. If its common ratio is 2, then its first term is
(a) 1
(b) 3
(c) 8
(d) none of these

Answer:

an=128, Sn=225 and r=2an=128 arn-1=1282n-1a=1282na2=128 2n=256  a             ........(i)Also, Sn=225arn-1r-1=225a2n-12-1=225a256  a-1=225        Using (i)256-a=225a=256-225a=31

Disclaimer: None of the given options are correct. This solution has been created according to the question given in the book.

Page No 20.57:

Question 13:

If second term of a G.P. is 2 and the sum of its infinite terms is 8, then its first term is
(a) 1/4
(b) 1/2
(c) 2
(d) 4

Answer:

(d) 4

 a2=2  ar=2            ........(i)Also, S=8a1-r=8a1-2a=8          Using (i)a2=8a-2a2-8a+16=0a-42=0a=4

Page No 20.57:

Question 14:

If a, b, c are in G.P. and x, y are AM's between a, b and b,c respectively, then

(a) 1x+1y=2

(b) 1x+1y=12

(c) 1x+1y=2a

(d) 1x+1y=2b

Answer:

(d) 1x+1y=2b

a, b and c are in G.P. b2=ac                   ........(i)a, x and b are in A.P. 2x=a+b              ........(ii)Also, b, y and c are in A.P. 2y=b+c         2y=b+b2a             Using (i)2y=b+b22x-b     Using (ii)2y=b2x-b+b22x-b2y=2bx-b2+b22x-b2y=2bx2x-by=bx2x-by2x-b=bx2xy-by=bxbx+by=2xyDividing both the sides by xy:1y+1x=2b



Page No 20.58:

Question 15:

If A be one A.M. and p, q be two G.M.'s between two numbers, then 2 A is equal to
(a) p3+q3pq

(b) p3-q3pq

(c) p2+q22

(d) pq2

Answer:

(a) p3+q3pq

Let the two positive numbers be a and b. a, A and b are in A.P. 2A=a+b                  (i)Also, a, p, q and b are in G.P. r=ba13Again, p=ar and q=ar2.          (ii)Now, 2A=a+b       From (i)=a+aba=a+aba133=a+ar3=ar2ar2+ar22ar=p2q+q2p              Using (ii)=p3+q3pq

Page No 20.58:

Question 16:

If p, q be two A.M.'s and G be one G.M. between two numbers, then G2 =
(a) (2p q) (p − 2q)
(b) (2pq) (2qp)
(c) (2pq) (p + 2q)
(d) none of these

Answer:

(a) (2p q) (p − 2q)

Let the two numbers be a and b.a, p, q and b are in A.P. p-a=q-p=b-q                    p-a=q-p  and  q-p=b-qa=2p-q  and  b= 2q-p        (i)Also, a, G and b are in G.P. G2=abG2=2p-q2q-p

Page No 20.58:

Question 17:

If x is positive, the sum to infinity of the series
11+x-1-x(1+x)2+(1-x)2(1+x)3-(1-x)3(1+x)4+...... is
(a) 1/2
(b) 3/4
(c) 1
(d) none of these

Answer:

(a) 12

Let S=11+x-1-x1+x2+1-x21+x3-1-x31+x4+ ... It is clear that it is a G.P. with a=11+x and r=-1-x1+x. S=a1-rS=11+x1--1-x1+xS=11+x1+1-x1+xS=11+x1+x+1-x1+xS=12

Page No 20.58:

Question 18:

If x = (43) (46) (46) (49) .... (43x) = (0.0625)−54, the value of x is
(a) 7
(b) 8
(c) 9
(d) 10

Answer:

(b) 8

 434649412...43x=0.0625-5443+6+9+12+ ... +3x=62510000-54431+2+3+4+ ... +x=116-5443xx+12=116-5443xx+12=4-2-54Comparing both the sides:3xx+12=108xx+1=72x2+x-72=0x2+9x-8x-72=0xx+9-8x+9=0x+9x-8=0x=8, -9x=8    [ x is positive]

Page No 20.58:

Question 19:

Given that x > 0, the sum n=1xx+1n-1 equals
(a) x

(b) x + 1

(c) x2x+1

(d) x+12x+1

Answer:

(b) x + 1

n=1xx+1n-1=1+xx+1+xx+12+xx+13+xx+14+ ... =11-xx+1               it is a G.P. with a =1 and r=xx+1=x+1x+1-x=x+11=x+1

Page No 20.58:

Question 20:

In a G.P. of even number of terms, the sum of all terms is five times the sum of the odd terms. The common ratio of the G.P. is
(a) -45
(b) 15
(c) 4
(d) none of these

Answer:

(c) 4

Let there be 2n terms in a G.P.Let a be the first term and r be the common ratio. S2n=5Sodd termsar2n-1r-1=5a+ar2+ar4+ar6+ ... ar2n-1ar2n-1r-1=5ar2n-1r2-1r2n-1r-1=5r2n-1r2-1rn2-12r-1=5rn2-12r2-1rn-1rn+1r-1=5rn-1rn+1r-1r+1rn-1rn+1r-1r+1-5r-1rn-1rn+1=0rn-1rn+1r-1r+1-5=0But, r =1 or -1 is not possible. r=4

Page No 20.58:

Question 21:

Let x be the A.M. and y, z be two G.M.s between two positive numbers. Then, y3+z3xyz is equal
to
(a) 1
(b) 2
(c) 12
(d) none of these

Answer:

(b) 2

Let the two numbers be a and b.a, x and b are in A.P. 2x=a+b                                       (i)Also, a, y, z and b are in G.P. ya=zy=bzy2=az , yz=ab, z2=by            (ii)Now, y3+z3xyz=y2xz+z2xy                  =1xy2z+z2y=1xazz+byy                          Using (ii)=1xa+b=2a+ba+b                            Using (i)=2

Page No 20.58:

Question 22:

The product (32), (32)1/6 (32)1/36 ... to ∞ is equal to
(a) 64
(b) 16
(c) 32
(d) 0

Answer:

(a) 64

32×3216×32136× ... =321+16+136+ ... = 3211-16   [ it is a G.P.]=3265=2565=26=64

Page No 20.58:

Question 23:

The two geometric means between the numbers 1 and 64 are
(a) 1 and 64
(b) 4 and 16
(c) 2 and 16
(d) 8 and 16
(e) 3 and 16

Answer:

(b) 4 and 16

Let the two G.M.s between 1 and 64 be G1 and G2.Thus, 1, G1, G2 and 64 are in G.P.  64=1×r3r=643r=4G1=ar=1×4=4And, G2=ar2=1×42=16Thus, 4 and 16 are the required G.M.s.

Page No 20.58:

Question 24:

In a G.P. if the (m + n)th term is p and (mn)th term is q, then its mth term is
(a) 0
(b) pq
(c) pq
(d) 12(p+q)

Answer:

(c) pq

Here, am+n=parm+n-1=p              .......(i)Also, am-n=qarm-n-1=q             .......(ii)Mutliplying (i) and (ii):arm+n-1arm-n-1=pqa2r2m-2=pqarm-12=pqarm-1=pqam=pqThus, the mth term is pq.

Page No 20.58:

Question 25:

Mark the correct alternative in the following question:

Let S be the sum, P be the product and R be the sum of the reciprocals of 3 terms of a G.P. Then p2R3 : S3 is equal to

(a) 1 : 1                        (b) (Common ratio)n : 1                        (c) (First term)2 : (Common ratio)2                        (d) None of these

Answer:

Let the three terms of the G.P. be ar, a, ar. ThenS=ar+a+ar=a1r+1+r=a1+r+r2r=ar2+r+1rAlso,P=ar×a×ar=a3And,R=ra+1a+1ar=1ar+1+1r=1ar2+r+1rNow,P2R3S3=a32×1ar2+r+1r3ar2+r+1r3=a6×1a3r2+r+1r3a3r2+r+1r3=11So, the ratio is 1:1.

Hence, the correct alternative is option (a).



Page No 20.9:

Question 1:

Show that each one of the following progressions is a G.P. Also, find the common ratio in each case:

(i) 4, −2, 1, −1/2, ...

(ii) −2/3, −6, −54, ...

(iii) a,3a24,9a316, ...

(iv) 1/2, 1/3, 2/9, 4/27, ...

Answer:

i We have,a1= 4, a2 =-2, a3 =1, a4 =-12Now, a2a1=-24=-12, a3a2=1-2, a4a3=-121=-12  a2a1=a3a2=a4a3=-12Thus, a1, a2, a3 and a4 are in G.P., where a=4 and r=-12.

ii We have,a1=-23 , a2 =-6, a3 =-54Now, a2a1=-6-23=9, a3a2=-54-6=9            a2a1=a3a2=9Thus, a1, a2  and a3 are in G.P., where a=-23 and r=9.

iii We have,a1=a , a2 =3a24, a3 =9a316Now, a2a1=3a24a=3a4, a3a2=9a3163a24=3a4            a2a1=a3a2=3a4Thus, a1, a2  and a3 are in G.P., where the first term is a and the common ratio is 3a4.

iv We have,a1=12 , a2 =13, a3 =29, a4 =427Now, a2a1=1312=23, a3a2=2913=23, a4a3= 42729=23 a2a1=a3a2=a4a3=23Thus, a1, a2, a3 and a4  are in G.P., where the first term is 12 and the common ratio is 23.



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