The sum of first n terms of two APs are in teh ratio (3n +8) : (7n + 15).
Find hte ratio of their 12th terms.

Open in App

Solution

Let $a_{1}, a_{2}$ be the first terms and $d_{1}, d_{2}$ be the common difference of the two A.Ps.

And $S_{n}, S_{n}^{'}$ be the sum of their n terms.

Sum of n terms is given by,

$S_{n} = \frac{n}{2} [2 a_{1} + (n - 1) d_{1}]$

$S_{n}^{'} = \frac{n}{2} [2 a_{2} + (n - 1) d_{2}]$

$\frac{S_{n}}{S_{n}^{'}} = \frac{\frac{n}{2} [2 a_{1} + (n - 1) d_{1}]}{\frac{n}{2} [2 a_{2} + (n - 1) d_{2}]} = \frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}}$

$\Rightarrow \frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}} = \frac{3 n + 8}{7 n + 15}$ ----(1)

$\Rightarrow \frac{a_{1} + \frac{(n - 1)}{2} d_{1}}{a_{2} + \frac{(n - 1)}{2} d_{2}} = \frac{3 n + 8}{7 n + 15}$

12th term of an AP is given by a + 11d,

therefore on taking $\frac{(n - 1)}{2} = 11 n - 1 = 22 n = 23$

$\frac{a_{12}}{a_{12}^{'}} = \frac{a_{1} + 11 d_{1}}{a_{2} + 11 d_{2}} = \frac{3 \times 23 + 8}{7 \times 23 + 15} \frac{a_{12}}{a_{12}^{'}} = \frac{77}{176}$

Thus, the ratio of 12th terms of the two A.Ps is 77:176.

Suggest Corrections

Similar questions

n

A . P^{'} s

(3 n + 8) : (7 n + 15)

12^{t h}

n

(3 n + 8) : (7 n + 15)

12^{t h}

3 n + 8 : 7 n + 15

12

n

2

(3 n + 8) : (7 n + 15)

12^{t h}

Join BYJU'S Learning Program