# The Sums of First N Terms of Two A.P.'S Are in the Ratio (7n + 2) : (N + 4). Find the Ratio of Their 5th Terms. - Mathematics

The sums of first n terms of two A.P.'s are in the ratio (7n + 2) : (n + 4). Find the ratio of their 5th terms.

#### Solution

$\text { Let the first term, the common difference and the sum of the first n terms of the first A . P . be} a_1 , d_1 and S_1 , respectively, and those of the second A . P . be a_2 , d_2 and S_2 , respectively .$

$\text { Then, we have },$

$S_1 = \frac{n}{2}\left[ 2 a_1 + \left( n - 1 \right) d_1 \right]$

$\text { And, } S_2 = \frac{n}{2}\left[ 2 a_2 + \left( n - 1 \right) d_2 \right]$

$\text { Given: }$

$\frac{S_1}{S_2} = \frac{\frac{n}{2}\left[ 2 a_1 + \left( n - 1 \right) d_1 \right]}{\frac{n}{2}\left[ 2 a_2 + \left( n - 1 \right) d_2 \right]} = \frac{7n + 2}{n + 4}$

$\Rightarrow \frac{S_1}{S_2} = \frac{\left[ 2 a_1 + \left( n - 1 \right) d_1 \right]}{\left[ 2 a_2 + \left( n - 1 \right) d_2 \right]} = \frac{7n + 2}{n + 4}$

$\text { To find the ratio of the 5th terms of the two A . P . s, we replace n by } (2 \times 5 - 1 = 9)\text { in the above equation }:$

$\Rightarrow \frac{\left[ 2 a_1 + \left( 9 - 1 \right) d_1 \right]}{\left[ 2 a_2 + \left( 9 - 1 \right) d_2 \right]} = \frac{7 \times 9 + 2}{9 + 4}$

$\Rightarrow \frac{\left[ 2 a_1 + \left( 8 \right) d_1 \right]}{\left[ 2 a_2 + \left( 8 \right) d_2 \right]} = \frac{7 \times 9 + 2}{9 + 4} = \frac{65}{13}$

$\Rightarrow \frac{\left[ a_1 + 4 d_1 \right]}{\left[ a_2 + 4 d_2 \right]} = \frac{5}{1} = 5: 1$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 19 Arithmetic Progression
Exercise 19.4 | Q 34 | Page 31