Let a 1- a 2- a 3- - be the terms of an AP- If a 1- a 2- a 3- a p - a 1- a 2- a 3- a q - p 2- q 2- p - q- find the value of a 6- a 21-

Given, a_1,a_2,a_3,... are in AP. and a_1+a_2+a_3+...+a_p/a_1+a_2+a_3+...+a_q=p^2/q^2 ⇒ p/2[2a_1+(p 1)d ]/q/2[2a_1+(q 1)d ]=p^2/q^2 [∵ S_n=n/2{ 2a+(n 1)d ...

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Byju's Answer

Standard XI

Mathematics

Geometric Progression

Let a 1, a 2,...

Question

Let $a_{1}, a_{2}, a_{3}, . . .$ be the terms of an AP. If $\frac{a_{1} + a_{2} + a_{3} + . . . + a_{p}}{a_{1} + a_{2} + a_{3} + . . . + a_{q}} = \frac{p^{2}}{q^{2}}, p \neq q,$ find the value of $\frac{a_{6}}{a_{21}}$ .

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Solution

Given, $a_{1}, a_{2}, a_{3}, . . .$ are in AP.

and $\frac{a_{1} + a_{2} + a_{3} + . . . + a_{p}}{a_{1} + a_{2} + a_{3} + . . . + a_{q}} = \frac{p^{2}}{q^{2}}$

$\Rightarrow \frac{p / 2 [2 a_{1} + (p - 1) d]}{q / 2 [2 a_{1} + (q - 1) d]} = \frac{p^{2}}{q^{2}}$

$[∵ S_{n} = \frac{n}{2} {2 a + (n - 1) d}]$

$\Rightarrow \frac{2 a_{1} + (p - 1) d}{2 a_{1} + (q - 1) d} = \frac{p}{q}$

$\Rightarrow 2 a_{1} q + p q d - q d = 2 a_{1} p + p q d - p d$

$\Rightarrow 2 a_{1} q - q d - 2 a_{1} p + p d = 0$

$\Rightarrow 2 a_{1} p - 2 a_{1} q - p d + q d = 0$

$\Rightarrow 2 a_{1} (p - q) - d (p - q) = 0$

$\Rightarrow (2 a_{1} - d) (p - q) = 0$

$\Rightarrow 2 a_{1} - d = 0 [∵ p \neq q]$

$\Rightarrow d = 2 a_{1}$

Now, $\frac{a_{6}}{a_{21}} = \frac{a_{1} + 5 d}{a_{1} + 20 d} = \frac{a_{1} + 10 a_{1}}{a_{1} + 40 a_{1}} = \frac{11}{41}$

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Let a1,a2,a3,... be the terms of an AP. If a1+a2+a3+...+apa1+a2+a3+...+aq=p2q2, p≠q, find the value of a6a21.

Let $a_{1}, a_{2}, a_{3}, . . .$ be the terms of an AP. If $\frac{a_{1} + a_{2} + a_{3} + . . . + a_{p}}{a_{1} + a_{2} + a_{3} + . . . + a_{q}} = \frac{p^{2}}{q^{2}}, p \neq q,$ find the value of $\frac{a_{6}}{a_{21}}$ .