Math Ncert Exemplar 2019 Solutions for Class 11 Science Maths Chapter 9 - Sequence And Series

Answer:

Let the common difference of an AP is d.
According to the question.
S_p = 0

$$\begin{gathered} \Rightarrow \frac{p}{2}\left[2a+\left(p-1\right)d\right]=0 \\ \Rightarrow 2a+\left(p-1\right)d=0 \\ \therefore d=\frac{-2a}{p-1} \end{gathered}$$

Now, sum of next q terms = S_p+q − S_p = S_p+q − 0

$$\begin{gathered} =\frac{p+q}{2}\left[2a+\left(p+q-1\right)d\right] \\ =\frac{p+q}{2}\left[2a+\left(p+q-1\right)d\right] \\ =\frac{p+q}{2}\left[2a+\left(p-1\right)d+qd\right] \\ =\frac{p+q}{2}\left[2a+\left(p-1\right)\times \frac{-2a}{p-1}+q\times \frac{-2a}{p-1}\right] \\ =\frac{p+q}{2}\left[2a-2a-\frac{2aq}{p-1}\right] \\ =\frac{p+q}{2}\left[-\frac{2aq}{p-1}\right] \\ =-\frac{a\left(p+q\right)q}{p-1} \end{gathered}$$



 

Answer:

Let us assume that the man saved Rs a in the first year.

Answer:

Given: Salary per month = Rs. 5200

Increased per month = Rs. 320

(a) Salary for the 10th month = 5200 + (10 − 1)320

= 5200 + 2880

= 8080

Hence, the salary for 10th month will be Rs. 8080.

(b) Total earnings during the first year is equal to the sum of 12 term of the AP.

Total earning = S₁₂
$$\begin{gathered} =\frac{12}{2}\left[2\times 5200+\left(12-1\right)320\right] \\ =6\left[10400+11\times 320\right] \\ =6\left[10400+3520\right] \\ =6\left[13920\right] \\ =83520 \end{gathered}$$

Hence, the salary for first year is Rs. 83520.

Answer:

Let the first term and common ratio of GP be a and r, respectively.

$$\begin{gathered} \mathrm{Given} \mathrm{that}, p^{\mathrm{th}} term=q\Rightarrow a{r}^{p-1}=q .....\left(1\right) \\ and q^{\mathrm{th}} term=p\Rightarrow a{r}^{q-1}=p .....\left(2\right) \end{gathered}$$

On dividing (1) by (2), we get

$$\begin{gathered} \frac{a{r}^{p-1}}{a{r}^{q-1}}=\frac{q}{p} \\ \Rightarrow r^{p-q}=\frac{q}{p} \\ \Rightarrow r={\left(\frac{q}{p}\right)}^{\frac{1}{p-q}} \end{gathered}$$

Put the value of r in (1), we get

$$\begin{gathered} a{\left(\frac{q}{p}\right)}^{\frac{p-1}{p-q}}=q \\ \Rightarrow a=q{\left(\frac{p}{q}\right)}^{\frac{p-1}{p-q}} \end{gathered}$$

(p + q)^th term, T_p+q = ar ^{p + q − 1}

$$\begin{gathered} =q{\left(\frac{p}{q}\right)}^{\frac{p-1}{p-q}}{\left(\frac{q}{p}\right)}^{\frac{p+q-1}{p-q}} \\ =q{\left(\frac{p}{q}\right)}^{\frac{p-1}{p-q}-\frac{p+q-1}{p-q}} \\ =q{\left(\frac{q}{p}\right)}^{\frac{q}{p-q}} \\ =\left(\frac{q^{{\displaystyle \frac{q}{p-q}}+1}}{p^{{\displaystyle \frac{q}{p-q}}}}\right) \\ ={\left(\frac{q^p}{p^q}\right)}^{\frac{1}{p-q}} \end{gathered}$$

Answer:

Here, a = 5 and d = 2

Let the carpenter finish the job in n days.

Then, S_n = 192

$$\begin{gathered} \Rightarrow 192=\frac{n}{2}\left[2a+\left(n-1\right)d\right] \\ \Rightarrow 192=\frac{n}{2}\left[2\times 5+\left(n-1\right)2\right] \\ \Rightarrow 192=n\left[5+n-1\right] \\ \Rightarrow n^2+4n-192=0 \\ \Rightarrow \left(n-12\right)\left(n+16\right)=0 \\ \Rightarrow n=12 \end{gathered}$$

Hence, it take him 12 days to finish the job.

Answer:

We know that, sum of interior angles of a polygon of side n is (n − 2) × 180°.

Let t_n = (n − 2) × 180°
Since t_n is linear in n, it is n^th term of some AP.
 t₃ = a = (3 − 2) × 180° = 180°
Common difference, d = 180°

Thus, the AP is 180°, 360°, 540°, ....

Sum of the interior angle for a 21 sided polygon is 

t₂₁ = (21 − 2) × 180° = 3420°
 

Answer:

Let the given equilateral triangle be ∆ ABC with each side of 20 cm.
By joining the mid-points of this triangle, we get another equilateral triangle of side equal to half of the length of side of ∆ABC.
Continuing in this way, we get a set of equilateral triangles with side equal to half of the side of the previous triangle.
Now,
Perimeter of first triangle = 20 x 3 = 60 cm;
Perimeter of second triangle = 10 x 3 = 30 cm;
Perimeter of third triangle = 5×3 = 15 cm;

Clearly, 60, 30, 15, ... form a GP with a = 60, and $r=\frac{30}{60}=\frac{1}{2}$

We have to find the perimeter of the sixth inscribed triangle i.e., we have to find the sixth term of the GP.

$$\begin{gathered} \therefore \mathrm{Perimeter} \mathrm{of} \mathrm{sixth} \mathrm{inscribed} \mathrm{triangle}=a_6 \\ =a{r}^{6-1} \\ =60\times {\left(\frac{1}{2}\right)}^5 \\ =\frac{60}{32} \\ =\frac{15}{8} \mathrm{cm} \end{gathered}$$ 

Answer:

Distance travelled to bring first potato = 24 + 24 = 2 x 24 = 48 m
Distance travelled to bring second potato = 2(24 + 4) = 2 x 28 = 56 m
Distance travelled to bring third potato = 2(24 + 4 + 4) = 2 X 32 = 64 m; and so on…
Clearly, 48, 56, 64,… is an A.P. with first term 48 and common difference 8. Also, number of terms is 20.
Total distance run in bringing back all the potatoes,

$\begin{array}{rcl}{S}_{20}& =& \frac{20}{2}\left[2\times 48+\left(20-1\right)8\right]\\ & =& 20\left[48+76\right]\\ & =& 20\times 124\\ & =& 2480 \mathrm{m}\end{array}$

Answer:

Let the first place team get Rs. a as the prize money.
Since award money increases by the same amount for successive finishing places, we get an AP.

Let the constant amount be d.

$$\begin{gathered} \mathrm{Here}, t_{16}=275, n=16, \mathrm{and} S_{16}=8000 \\ \therefore t_{16}=a+\left(16-1\right)\left(-d\right) \\ \Rightarrow 275=a-15d .....\left(1\right) \\ Also, S_{16}=\frac{16}{2}\left[2a+\left(n-1\right)\left(-d\right)\right] \\ \Rightarrow 8000=8\left[2a-15d\right] \\ \Rightarrow 1000=2a-15d .....\left(2\right) \\ \mathrm{From} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get} \\ a=725 \end{gathered}$$

Hence, first place team receives â‚¹725.
 

Answer:

Given that, a₁, a₂, ..., a_n are in AP, for all a_i > 0.

∴ a₁ − a₂ = a₂ − a₃ = ... = a_{n − 1} − a_n = −d (contant)
$$\begin{gathered} \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+...+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}} \\ =\frac{\sqrt{a_1}-\sqrt{a_2}}{a_1-a_2}+\frac{\sqrt{a_2}-\sqrt{a_3}}{a_2-a_3}+...+\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{a_{n-1}-a_n} \\ =\frac{\sqrt{a_1}-\sqrt{a_2}}{-d}+\frac{\sqrt{a_2}-\sqrt{a_3}}{-d}+...+\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{-d} \\ =\frac{1}{-d}\left[\sqrt{a_1}-\sqrt{a_n}\right] \\ =\frac{a_1-a_n}{-d\left(\sqrt{a_1}+\sqrt{a_n}\right)} \\ =\frac{-\left(n-1\right)d}{-d\left(\sqrt{a_1}+\sqrt{a_n}\right)} \left[\mathrm{as} a_n=a_1+\left(n-1\right)d\right] \\ =\frac{\left(n-1\right)}{\sqrt{a_1}+\sqrt{a_n}} \end{gathered}$$

Answer:

Given series is: (3³ – 2³) + (5³ – 4³) + (7³ – 6³) + ... n terms

$\begin{array}{rcl}{T}_n& =& {\left(2n+1\right)}^3-{\left(2n\right)}^3\\ & =& \left(2n+1-2n\right)\left[{\left(2n+1\right)}^2+\left(2n+1\right)2n+{\left(2n\right)}^2\right]\\ & =& 12n^2+6n+1\end{array}$

(i) Sum of n terms,

$\begin{array}{rcl}{S}_n& =& \sum _{n=1}^n\left(12n^2+6n+1\right)\\ & =& 12·\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{6n\left(n+1\right)}{2}+2\\ & =& 2n\left(n+1\right)\left(2n+1\right)+6n\left(n+1\right)+2\\ & =& 4n^3+9n^2+6n\end{array}$

(ii) Sum of 10 terms, S₁₀ = 4 × (10)³ + 9 × (10)² + 6 × 10 = 4960

Answer:

Given: Sum of whose first n terms is 2n + 3n².

$\begin{array}{rcl}\therefore T_n& =& S_n-S_{n-1}\\ & =& \left(2n+3n^2\right)-\left[2\left(n-1\right)+3{\left(n-1\right)}^2\right]\\ & =& 2+3\left(n-n+1\right)\left(n+n-1\right)\\ & =& 2+3\left(2n-1\right)\\ & =& 6n-1\\ \therefore T_r& =& 6r-1\end{array}$
 

Answer:

Let the number be a and b.

$$\begin{gathered} \mathrm{Then}, A=\frac{a+b}{2} \mathrm{or} 2A=a+b \\ \mathrm{Also}, G_1 \mathrm{and} G_2 \mathrm{are} \mathrm{geometric} \mathrm{means} \mathrm{between} a \mathrm{and} b, \mathrm{then} a, G_1, G_2, b \mathrm{are} \mathrm{in} \mathrm{GP}. \\ \mathrm{Let} r \mathrm{be} \mathrm{the} \mathrm{common} \mathrm{ratio}. \\ \mathrm{Then}, b=a{r}^{4-1}=a{r}^3 \\ \Rightarrow \frac{b}{a}=r^3 \\ \Rightarrow r={\left(\frac{b}{a}\right)}^{\frac{1}{3}} \\ \therefore G_1=ar=a{\left(\frac{b}{a}\right)}^{\frac{1}{3}}=a^{\frac{2}{3}}{b}^{\frac{1}{3}} \\ \mathrm{and} G_2=a{r}^2=a{\left(\frac{b}{a}\right)}^{\frac{2}{3}}=a^{\frac{1}{3}}{b}^{\frac{2}{3}} \end{gathered}$$

$\begin{array}{rcl}\therefore \frac{{G_1}^2}{G_2}+\frac{{G_2}^2}{G_1}& =& \frac{{G_1}^3+{G_2}^3}{G_1{G}_2}\\ & =& \frac{a^{2}b+a{b}^2}{ab}\\ & =& a+b\\ & =& 2A\end{array}$
 

Answer:

Since θ₁, θ₂, θ₃, ..., θ_n are in AP, we get
θ₂ − θ₁ = θ₃ − θ₂ = ... = θ_n − θ_n-1 = d
Now, 

$$\begin{gathered} \sec {\theta }_1\sec {\theta }_2 \\ =\frac{1}{\sin d}·\frac{\sin d}{\cos {\theta }_1\cos {\theta }_2} \\ =\frac{1}{\sin d}·\frac{\sin \left({\theta }_2-{\theta }_1\right)}{\cos {\theta }_1\cos {\theta }_2} \\ =\frac{1}{\sin d}·\frac{\left(\sin {\theta }_2\cos {\theta }_1-\cos {\theta }_2\sin {\theta }_1\right)}{\cos {\theta }_1\cos {\theta }_2} \\ =\frac{\tan {\theta }_2-\tan {\theta }_1}{\sin d} \end{gathered}$$

$$\begin{gathered} \mathrm{Similarly}, \sec {\theta }_2\sec {\theta }_3=\frac{\tan {\theta }_3-\tan {\theta }_2}{\sin d} \\ \sec {\theta }_3\sec {\theta }_4=\frac{\tan {\theta }_4-\tan {\theta }_3}{\sin d} \\ \therefore s\mathrm{ec}{\theta }_1\sec {\theta }_2+s\mathrm{ec}{\theta }_2\sec {\theta }_3+...+s\mathrm{ec}{\theta }_{n-1}\sec {\theta }_n=\frac{\tan {\theta }_n-\tan {\theta }_{n-1}}{\sin d} \end{gathered}$$
 

Answer:

Let first term and common difference of the A.P. be a and d, respectively. Given, S_p = q.



$$\begin{gathered} \Rightarrow \frac{p}{2}\left[2a+\left(p-1\right)d\right]=q\Rightarrow 2a+\left(p-1\right)d=\frac{2q}{p} .....\left(1\right) \\ \mathrm{Also}, S_q=p \\ \Rightarrow \frac{q}{2}\left[2a+\left(q-1\right)d\right]=p\Rightarrow 2a+\left(p-1\right)d=\frac{2p}{q} .....\left(2\right) \\ \mathrm{On} \mathrm{subracting} \left(2\right) \mathrm{from} \left(1\right), \mathrm{we} \mathrm{get} \\ \left(p-q\right)d=\frac{2q}{p}-\frac{2p}{q} \\ \Rightarrow \left(p-q\right)d=\frac{2\left(q^2-p^2\right)}{pq} \\ \Rightarrow d=\frac{-2\left(p+q\right)}{pq} .....\left(3\right) \\ \mathrm{Put} \mathrm{the} \mathrm{value} \mathrm{of} d \mathrm{in} \left(1\right), \mathrm{we} \mathrm{get} \\ 2a+\left(p-1\right)\left(\frac{-2\left(p+q\right)}{pq}\right)=\frac{2q}{p} \\ \Rightarrow a=\frac{q}{p}+\frac{\left(p+q\right)\left(p-1\right)}{pq} .....\left(4\right) \\ \mathrm{Now}, S_{p+q}=\frac{p+q}{2}\left[2a+\left(p+q-1\right)d\right] \\ =\frac{p+q}{2}\left[\frac{2q}{p}+\frac{2\left(p+q\right)\left(p-1\right)}{pq}-\frac{2\left(p+q-1\right)\left(p+q\right)}{pd}\right] \\ =\left(p+q\right)\left[\frac{q}{p}+\frac{\left(p+q\right)\left(p-1-p-q+1\right)}{pq}\right] \\ =\left(p+q\right)\left[\frac{q}{p}+\frac{\left(p+q\right)\left(-q\right)}{pq}\right] \\ =\left(p+q\right)\left[\frac{q}{p}-\frac{p+q}{p}\right] \\ =-\left(p+q\right) \\ \mathrm{Also}, \mathrm{Now}, S_{p-q}=\frac{p-q}{2}\left[2a+\left(p-q-1\right)d\right] \\ =\frac{p-q}{2}\left[\frac{2q}{p}+\frac{2\left(p+q\right)\left(p-1\right)}{pq}-\frac{2\left(p-q-1\right)\left(p+q\right)}{pd}\right] \\ =\left(p-q\right)\left[\frac{q}{p}+\frac{\left(p+q\right)\left(p-1-p+q+1\right)}{pq}\right] \\ =\left(p-q\right)\left[\frac{q}{p}+\frac{\left(p+q\right)\left(q\right)}{pq}\right] \\ =\left(p-q\right)\left[\frac{q}{p}+\frac{p+q}{p}\right] \\ =\left(p-q\right)\left(\frac{p+2q}{p}\right) \end{gathered}$$

Answer:

Let A and d be the first term and common difference of AP, respectively. Also, let B and R be the first term and common ratio of GP, respectively.
It is given that,

A + (p − 1) = a          .....(1)
A + (q − 1) = b          .....(2)
A + (r − 1) = c           .....(3)

Also, a = BR^p − 1             .....(4)
b = BR^q − 1             .....(5)
c = BR^r − 1             .....(6)

Subtracting (2) from (1), we get

a − b = d(p − q)

Subtracting (3) from (2), we get

b − c = d(q − r)

Subtracting (1) from (3), we get

c − a = d(r − p)

a^b–c . b^c – a . c^a – b = (BR^(p − 1))^{d(q – r)} . (BR^(q − 1))^d(r – p) . (BR^(r − 1))^{d(p – q)}
= B^{d[(q − r) + (r − p) + (p − q)]}R^{d[(p − 1)(q − r) + (q − 1)(r − p) + (r − 1)(p − q)]}
=B⁰R⁰ = 1

Answer:

Given: S_n = 3n + 2n²

S₁ = 3(1) + 2(1)² = 5

S₂ = 3(2) + 2(2)² = 10

S₂ − S₁ = 9 = t₂

d = t₂ − t₁ = 9 − 5 = 4

Hence, the correct answer is option (D).

Answer:

Let a and r be the first term and common ration, respectively.

Given that the third term is 4.

∴ ar² = 4

Product of first 5 terms = a × ar × ar² × ar³ × ar⁴ = a⁵r¹⁰ = (ar²)⁵ = (4)⁵

Hence, the correct answer is option (B).

Answer:

Let the first term and common difference of given A.P. be a and d, respectively.

It is given that

9 × t₉ = 13 × t₁₃

⇒ 9(a + 8d) = 13(a + 12d)

⇒ 9a + 72d = 13a + 156d

⇒ 4a + 84d = 0

⇒ 4(a + 21d) = 0

⇒ t₂₂ = 0

Hence, the correct answer is option (A).

Answer:

Given: x, 2y, 3z are in AP. Therefore, 

$$\begin{gathered} 2y=\frac{\mathrm{x}+3\mathrm{z}}{2} \\ \Rightarrow 4\mathrm{y}=\mathrm{x}+3\mathrm{z} \end{gathered}$$

Also, x, y, z are in GP. Therefore

y = xr 

and z = xr² 
$$\begin{gathered} \therefore 4xr=x+3x{r}^2 \\ \Rightarrow 4r=1+3r^2 \\ \Rightarrow 3r^2-4r+1=0 \\ \Rightarrow \left(3r-1\right)\left(r-1\right)=0 \\ \Rightarrow r=\frac{1}{3} \left[\mathrm{For} r=1, x, y, z, \mathrm{are} \mathrm{same}\right] \end{gathered}$$

Hence, the correct answer is option (B).

Answer:

Given, S_n = qn² and S_m = qm²

∴ S₁ = q, S₂ = 4q, S₃ = 9q and S₄ = 16q

Now, t₁ = q

∴ t₂ = S₂ − S₁ = 4q − q = 3q

t₃ = S₃ − S₂ = 9q − 4q = 5q

t₄ = S₄ − S₃ = 16q − 9q = 7q

So, the AP is q, 3q, 5q, 7q, ....

Thus, first term is q and common difference is 3q − q = 2q

$\begin{array}{rcl}\therefore S_q& =& \frac{q}{2}\left[2q+\left(q-1\right)2q\right]\\ & =& \frac{q}{2}\left[2q+2q^2-2q\right]\\ & =& \frac{q}{2}\left[2q^2\right]\\ & =& q^3\end{array}$

Hence, the correct answer is option (C).
 

Answer:

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

Then, S_2n = 3S_n

$$\begin{gathered} \Rightarrow \frac{2n}{2}\left[2a+\left(2n-1\right)d\right]=\frac{3n}{2}\left[2a+\left(n-1\right)d\right] \\ \Rightarrow 2\left[2a+\left(2n-1\right)d\right]=3\left[2a+\left(n-1\right)d\right] \\ \Rightarrow 4a+\left(4n-2\right)d=6a+\left(3n-3\right)d \\ \Rightarrow 2a=\left(n+1\right)d \\ Now, \frac{S_{3n}}{S_n}=\frac{{\displaystyle \frac{3n}{2}}\left[2a+\left(3n-1\right)d\right]}{{\displaystyle \frac{2n}{2}}\left[2a+\left(n-1\right)d\right]} \\ =\frac{3\left[2a+\left(3n-1\right)d\right]}{2\left[2a+\left(n-1\right)d\right]} \\ =\frac{3\left[\left(n+1\right)d+\left(3n-1\right)d\right]}{2\left[\left(n+1\right)d+\left(n-1\right)d\right]} \\ =\frac{3}{2}\left(\frac{4nd}{nd}\right) \\ =6 \end{gathered}$$

Hence, the correct answer is option (B).

Answer:

 $$\begin{gathered} 4^x+4^{1-x}=4^x+\frac{4}{4^x} \\ ={\left(2^x-\frac{2}{2^x}\right)}^2+4 \\ \therefore 4^x+4^{1-x}\ge 4 \end{gathered}$$

Hence, the correct answer is option (B).

Answer:

$\begin{array}{rcl}\sum _{r=1}^n\frac{{\mathrm{S}}_r}{{\mathrm{S}}_r}& =& \sum _{r=1}^n\frac{{\left({\displaystyle \frac{r\left(r+1\right)}{2}}\right)}^2}{{\displaystyle \frac{r\left(r+1\right)}{2}}}\\ & =& \sum _{r=1}^n\frac{r\left(r+1\right)}{2}\\ & =& \frac{1}{2}\left[\sum _{r=1}^n{r}^2+\sum _{r=1}^{n}r\right]\\ & =& \frac{1}{2}\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}\right]\\ & =& \frac{1}{2}·\frac{n\left(n+1\right)}{2}\left[\frac{2n+1}{3}+1\right]\\ & =& \frac{n\left(n+1\right)}{4}\left[\frac{2n+4}{3}\right]\\ & =& \frac{n\left(n+1\right)\left(2n+1\right)}{6}\end{array}$

Hence, the correct answer is option (A).

Answer:

S₅₀ = 2 + 3 + 6 + 11 + 18 + ... + t₄₉ + t₅₀                 .....(1)

∴ S₅₀ = 0 + 3 + 6 + 11 + 18 + ... + t₄₉ + t₅₀             .....(2)

On subtracting (2) from (1), we get

0 = (2 + 1 + 3 + 5 + 7 + ...up tp 50 terms) − t₅₀

⇒ t₅₀ = 2 + [1 + 3 + 5 + 7 + ...upto 49 terms] 

$$\begin{gathered} =2+\frac{49}{2}\left[2\times 1+\left(49-1\right)\times 2\right] \\ =2+49\left(1+48\right) \\ =2+{\left(49\right)}^2 \end{gathered}$$

Hence, the correct answer is option (D).


 

Answer:

Let the length, breadth, and height of the rectangular solid block be $\frac{q}{r}, a, \mathrm{and} ar$ respectively.

$$\begin{gathered} \therefore \mathrm{Volume}=216 {\mathrm{cm}}^3 \\ \Rightarrow \frac{a}{r}\times a\times ar=216 \\ \Rightarrow a^3=216 \\ \Rightarrow a=6 \mathrm{cm} \end{gathered}$$

Also, Surface area = 

$$\begin{gathered} \Rightarrow 2\left(\frac{a}{r}\times a+a\times ar+ar\times \frac{a}{r}\right)=252 \\ \Rightarrow 2a^2\left(\frac{1}{r}+r+1\right)=252 \\ \Rightarrow 2\times 36\left(\frac{1+r^2+r}{r}\right)=252 \\ \Rightarrow 2\left(1+r^2+r\right)=7r \\ \Rightarrow 2r^2-5r+2=0 \\ \Rightarrow \left(2r-1\right)\left(r-2\right)=0 \\ \Rightarrow r=\frac{1}{2}, 2 \end{gathered}$$

For $r=\frac{1}{2}$, Length = 12 cm, Breadth = 6 cm and Height = 3 cm

For r = 2, Length = 3 cm, Breadth = 6 cm and Height = 12 cm
 

Answer:

Given: a, b, and c are in GP.

$\frac{b}{a}=\frac{c}{b}=r$

$\begin{array}{rcl}\frac{a-b}{b-c}& =& \frac{{\displaystyle \frac{a-b}{b}}}{{\displaystyle \frac{b-c}{b}}}\\ & =& \frac{{\displaystyle \frac{a}{b}}-1}{1-{\displaystyle \frac{c}{b}}}\\ & =& \frac{{\displaystyle \frac{1}{r}}-1}{1-r}\\ & =& \frac{{\displaystyle \frac{1-r}{r}}}{1-r}\\ & =& \frac{1}{r}\\ & =& \frac{a}{b}=\frac{b}{c}\end{array}$

 

Answer:

Let a be the first term and d be the common difference of the AP.

m^th term from the beginning is, a_m = a + (m − 1)d

and m^th term from the beginning is, a_n −m = a + (n − 1)d + (m − 1)d = a + (n − m)d

So, a_m + a_n −m = a + (m − 1)d + a + (n − m)d 

= a + a + (n − 1)d

= sum of first and last term of the AP.

Hence, â€‹the sum of terms equidistant from the beginning and end in an A.P. is equal to the sum of the first and last term of the AP.


 

Answer:

Let a and r be the first term and common ration, respectively.

Given that the third term is 4.

∴ ar² = 4

Product of first 5 terms = a × ar × ar² × ar³ × ar⁴ = a⁵r¹⁰ = (ar²)⁵ = (4)⁵

The third term of a G.P. is 4, the product of the first five terms is (4)⁵.

Answer:

False
Consider the sequence 4, 4, 4; which is A.P. and G.P. both.

Answer:

True; 
Consider the progression a, a + d, a + 2d, … and sequence of prime number 2, 3, 5, 7, 11,…
Clearly, progression is a sequence but sequence is not progression because it does not follow a specific pattern.

Answer:

True;

Consider any term a_r of an AP.

Now, a_{r + m} = a_r + (m − 1)d
And, a_{r − m} = a_r + (m − 1)(−d)

∴ a_r + m + a_r − m  = a_r + (m − 1)d + a_r + (m − 1)(−d)

⇒ a_r + m + a_r − m  = 2a_r

$\Rightarrow a_r=\frac{a_{r+m}+a_{r-m}}{2}$

Thus, any term of an AP (except first) is equal to half the sum of terms which are equidistant from it.
 

Answer:

Let two GPs be a, ar, ar², .... and b, bR, bR², .....

Sum of two GPs = (a + b), (ar + bR), (ar² + bR²), ....

Now, $\frac{ar+bR}{a+b}\ne \frac{a{r}^2+b{R}^2}{ar+bR}$

Similarly $\frac{ar-bR}{a-b}\ne \frac{a{r}^2-b{R}^2}{ar-bR}$

So, the sum or difference of two GPs are not a GP.

Answer:

False;

Sum of an AP is given by

$$\begin{gathered} S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right] \\ {S}_n=\left(\frac{2a-d}{2}\right)n+\frac{d}{2}{n}^2 \end{gathered}$$

Thus, S_n is of type an² + bn.

The general expression of a quadratic equation is of the form ax² + bx + c.

So, if the sum of n terms of a sequence is quadratic equation of the form ax² + bx + c, where c ≠ 0, then it does not represent sum of AP.

Answer:

Answer:

(a) The sum of squares of the first n natural numbers is given by 

(b) The sum of cubes of the first n natural numbers is given by 

(c) 2 + 4 + 6 + ... + 2n = 2( 1+ 2 + 3 + ... + n)$=2\left[\frac{n\left(n+1\right)}{2}\right]=n\left(n+1\right)$
(d) The sum of the first n natural numbers is given by 1 + 2 + 3 + 4 + … + n =