First slide

Arithmetic progression

Question

$Let a_{1}, a_{2} \dots a_{n} be the terms of an A.P. \frac{a_{1} + a_{2} + \dots a_{p}}{a_{1} + a_{2} + \dots + a_{1}} = \frac{p^{2}}{q^{2}}, p \neq q . T h e n \frac{a_{6}}{a_{21}} =$

Moderate

A

$\frac{11}{41}$
B

$\frac{11}{27}$
C

$\frac{12}{59}$
D

$\frac{16}{31}$

Solution

$\begin{array}{l} \frac{\frac{p}{2} [2 a_{1} + (p - 1) d]}{\frac{q}{2} [2 a_{1} + (q - 1) d]} = \frac{p^{2}}{q^{2}} \Rightarrow [\frac{2 a_{1} + (p - 1) d}{2 a_{1} + (q - 1) d}] = \frac{p}{q} \\ \Rightarrow \frac{a_{1} + (\frac{p - 1}{2}) d}{a_{1} + (\frac{q - 1}{2}) d} = \frac{p}{q} For \frac{a_{6}}{a_{21}}, p = 11, q = 41 \Rightarrow \frac{a_{6}}{a_{21}} = \frac{11}{41} \\ where \frac{p - 1}{2} = 5 p = 11 \\ \frac{q - 1}{2} = 20 q = 41 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App