Question

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Answer 1

Question:-

The sum of n term of two AP are in the ratio of

( 2n + 3):(5n -4). find the ratio of 10th term

Solution:-

$\rm \: S_n = \frac{n}{2} \{2a + (n - 1)d$

Let

$\rm \: ap_1 = a = a_1 \: and \: d = d_1$

or

$\rm \: ap_2 = a = a_2 \: and \: d = d_2$

We have to find

$\rm \frac{ a_1 + 9d_1}{a_2 + 9d_2} = {?}^{}$

Now

$\frac{ \frac{n}{2} \{2a_1 + (n - 1)d_1}{\frac{n}{2} \{2a_2 + (n - 1)d_2} = \frac{2n + 3}{5n - 4}$

Now multiply and divide by 2

$\frac{ 2a_1 + \frac{(n - 1)}{2} d_1}{2a_2 + \frac{(n - 1)}{2} d_2} = \frac{2n + 3}{5n - 4}$

$\frac{ a_1 + \frac{(n - 1)}{2} d_1}{a_2 + \frac{(n - 1)}{2} d_2} = \frac{2n + 3}{5n - 4}$

let n = 19

$\frac{ a_1 + \frac{(19 - 1)}{2} d_1}{a_2 + \frac{(19 - 1)}{2} d_2} = \frac{2 \times 19 + 3}{5 \times 19 - 4}$

$\frac{ a_1 + 9 d_1}{a_2 + 9d_2} = \frac{41}{91}$

So ratio of 10th term of AP is 41/91