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

Open in App

Solution

Let $A . P_{1}$ has first term $^{'} a_{1}^{'}$ & common difference $^{'} d_{1}^{'}$
$A . P_{2}$ has first term $^{'} a_{2}^{'}$ & common difference $^{'} d_{2}^{'}$
$s_{1} = \frac{n}{2} [2 a_{1} + (n - 1) d_{1}]$
$s_{2} = \frac{n}{2} [2 a_{2} + (n - 1) d_{2}]$
Ratio of their $12^{t h}$ terms
$\frac{a_{1} + (12 - 1) d_{1}}{a_{2} + (12 - 1) d_{2}} = \frac{a_{1} + 11 d_{1}}{a_{2} + 11 d_{2}} = \frac{77}{176} = \frac{7}{16}$
$\frac{s_{1}}{s_{2}} = \frac{2 a_{1} + (n - 1) d_{1}}{2 a_{1} + (n - 2) d - 2} = \frac{3 n + 8}{7 n + 15}$
Put $n = 23$
$\frac{s_{1}}{s_{2}} = \frac{2 a_{1} + 22 d_{1}}{2 a_{2} + 22 d_{2}} = \frac{3 (23) + 8}{7 (23) + 15}$
$\Rightarrow \frac{a_{1} + 11 d_{1}}{a_{2} + 11 d_{2}} = \frac{77}{166}$ .
Ratio is $77 : 166$

Suggest Corrections

Similar questions

n

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

(i) 12 t h

(i i) 15 t h

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

n

(7 n + 1) : (4 n + 17)

n

n

A . P^{'} s

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

12^{t h}

A . P

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

12 t h

Join BYJU'S Learning Program