First slide

Arithmetic progression

Question

If the sum of first $n$ terms of two A.P.’s are in the ratio $3 n + 8 : 7 n + 15,$ then the ratio of 12th term is

Moderate

A

$8 : 7$
B

$7 : 16$
C

$74 : 169$
D

$13 : 47$

Solution

Let two A.P.’s be
$a, a + d, a + 2 d, \dots$
and $A, A + D, A + 2 D, \dots$
We are given
$\begin{array}{l} \frac{\frac{n}{2} [2 a + (n - 1) d]}{\frac{n}{2} [2 A + (n - 1) D]} = \frac{3 n + 8}{7 n + 15} \\ \Rightarrow \frac{a + \frac{n - 1}{2} d}{A + \frac{n - 1}{2} D} = \frac{3 n + 8}{7 n + 15} \\ Put (n - 1) / 2 = 11 or n = 23 . \\ ∴ \frac{a + 11 d}{A + 11 D} = \frac{3 (23) + 8}{7 (23) + 15} = \frac{77}{176} = \frac{7}{16} \end{array}$

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