Let the terms of an arithmetic progression be | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $\frac{S_p}{S_q} = \frac{p^2}{q^2}$.Using the formula for the sum of $n$ terms of an $AP$,$S_n = \frac{n}{2}[2a_1 + (n-1)d]$,we have:$\frac{

Vedclass · Accepted Answer

$11/41$

Vedclass · Answer

(B) Given $\frac{S_p}{S_q} = \frac{p^2}{q^2}$.Using the formula for the sum of $n$ terms of an $AP$,$S_n = \frac{n}{2}[2a_1 + (n-1)d]$,we have:$\frac{\frac{p}{2}[2a_1 + (p-1)d]}{\frac{q}{2}[2a_1 + (q-1)d]} = \frac{p^2}{q^2}$$\frac{2a_1 + (p-1)d}{2a_1 + (q-1)d} = \frac{p}{q}$Dividing the numerator and denominator by $2$,we get:$\frac{a_1 + \frac{(p-1)}{2}d}{a_1 + \frac{(q-1)}{2}d} = \frac{p}{q} \dots (1)$We need to find $\frac{a_6}{a_{21}} = \frac{a_1 + 5d}{a_1 + 20d}$.Comparing this with equation $(1)$,we set $\frac{p-1}{2} = 5 \Rightarrow p = 11$ and $\frac{q-1}{2} = 20 \Rightarrow q = 41$.Substituting these values into equation $(1)$,we get $\frac{a_6}{a_{21}} = \frac{11}{41}$.

Let the terms of an arithmetic progression be $a_1, a_2, a_3, \dots$. If $\frac{a_1 + a_2 + \dots + a_p}{a_1 + a_2 + \dots + a_q} = \frac{p^2}{q^2}$,where $p \neq q$,then $\frac{a_6}{a_{21}} = \dots$

Similar Questions

The sum of $n$ terms of two arithmetic progressions are in the ratio $(3n + 8) : (7n + 15)$. Find the ratio of their $12^{\text{th}}$ terms.

If the sum of two extreme numbers of an $A.P.$ with four terms is $8$ and the product of the remaining two middle terms is $15$,then the greatest number of the series will be:

If the sides of a right-angled triangle are in $A.P.$,then their ratio is:

Given that $n$ arithmetic means are inserted between two sets of numbers $(a, 2b)$ and $(2a, b)$, where $a, b \in \mathbb{R}$. Suppose the $m^{th}$ means between these sets are equal, then the ratio $a : b$ is equal to:

Let $x, y > 0$. If $x^{3} y^{2} = 2^{15}$,then the least value of $3x + 2y$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module