Let a 1- a 2- a 3- - be terms of an A-P- If a 1- a 2- a p - a 1- a 2- a q - p 2- q 2- p - q then what is the value of a 6- a 21 - -3 MARKS-

Concept: 1 Mark Application: 2 Marks a_1, a_2, a_3, … be terms of AP ∴a_1 + a_2 + … + a_p/a_1 + a_2 + … + a_q = p^2/q^2 ⇒p/2[2a_1 + (p 1)d]/q/2[2a_1 + (q 1)d ...

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Byju's Answer

Standard X

Mathematics

Sum of 'n' Terms of an Arithmetic Progression

Let a 1, a 2,...

Question

Let $a_{1}, a_{2}, a_{3}, \dots \dots$ be terms of an A.P. If $\frac{a_{1} + a_{2} + \dots + a_{p}}{a_{1} + a_{2} + \dots + a_{q}} = \frac{p^{2}}{q^{2}}, p \neq q$ then what is the value of $\frac{a_{6}}{a_{21}}$ ? [3 MARKS]

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Solution

Concept: 1 Mark
Application: 2 Marks

$a_{1}, a_{2}, a_{3}, \dots$ be terms of AP

$∴ \frac{a_{1} + a_{2} + \dots + a_{p}}{a_{1} + a_{2} + \dots + a_{q}} = \frac{p^{2}}{q^{2}}$

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

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

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

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

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

$\Rightarrow 2 a_{1} = d$

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

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