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

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Solution

Let $a_{1}, a_{2}$ and $d_{1}, d_{2}$ be the first terms and common differences of the first and second AP respectively. Then, by the given condition, we have

$\frac{sum to n terms of first AP}{sum to n terms of second AP} = \frac{3 n + 8}{7 n + 15}$

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

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

Now, $\frac{12^{t h} term of first AP}{12^{t h} term of second AP} = \frac{a_{1} + 11 d_{1}}{a_{2} + 11 d_{2}}$

$= \frac{2 a_{1} + 22 d_{1}}{2 a_{2} + 22 d_{2}} = \frac{2 a_{1} + (23 - 1) d_{1}}{2 a_{2} + (23 - 1) d_{2}}$

$= \frac{3 \times 23 + 8}{7 \times 23 + 15}$ [putting n = 23 in (i)]

$= \frac{77}{176} = \frac{7}{16}$

Hence, the ratio of the 12th terms of given APs is 7 : 16.

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