If the ratio of the sum of n terms of two A.P | vedclass

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

Question

(C) Let $S_n$ and $S'_n$ be the sums of $n$ terms of two $A.P.'s$ with first terms $a, a'$ and common differences $d, d'$ respectively.The ratio of th

Vedclass · Accepted Answer

$4:3$

Vedclass · Answer

(C) Let $S_n$ and $S'_n$ be the sums of $n$ terms of two $A.P.'s$ with first terms $a, a'$ and common differences $d, d'$ respectively.The ratio of the sum of $n$ terms is given by:$\frac{S_n}{S'_n} = \frac{\frac{n}{2}[2a + (n - 1)d]}{\frac{n}{2}[2a' + (n - 1)d']} = \frac{7n + 1}{4n + 27}$$\frac{a + \frac{n - 1}{2}d}{a' + \frac{n - 1}{2}d'} = \frac{7n + 1}{4n + 27}$We want to find the ratio of the $11^{th}$ terms,which is $\frac{T_{11}}{T'_{11}} = \frac{a + 10d}{a' + 10d'}$.Comparing $\frac{n - 1}{2}$ with $10$,we get $n - 1 = 20$,so $n = 21$.Substituting $n = 21$ in the ratio:$\frac{T_{11}}{T'_{11}} = \frac{7(21) + 1}{4(21) + 27} = \frac{147 + 1}{84 + 27} = \frac{148}{111} = \frac{4}{3}$.Thus,the ratio is $4:3$.

If the ratio of the sum of $n$ terms of two $A.P.'s$ is $(7n + 1):(4n + 27)$,then the ratio of their $11^{th}$ terms is:

Similar Questions

The sum of the following series $1 + 6 + \frac{9(1^2 + 2^2 + 3^2)}{7} + \frac{12(1^2 + 2^2 + 3^2 + 4^2)}{9} + \frac{15(1^2 + 2^2 + ... + 5^2)}{11} + ...$ up to $15$ terms is:

If the $A.M.$ of two numbers is greater than $G.M.$ of the numbers by $2$ and the ratio of the numbers is $4:1$,then the numbers are

If the roots of the equation $x^3 - 9x^2 + \alpha x - 15 = 0$ are in $A.P.$,then $\alpha$ is:

The value of $\sum_{n=0}^{\infty} \frac{(n+1)^2}{7^n}$ is -

If $\frac{1}{p + q}, \frac{1}{r + p}, \frac{1}{q + r}$ are in $A.P.$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module