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Sum of N Terms of an AP

The sum of n ...

Question

The sum of n terms of two arithmetic progressions are in the ratio (3n + 8) : (7n + 15). What is the ratio of their 13th terms?

1 : 2

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94 : 178

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83 : 190

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2 : 3

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Solution

The correct option is C
83 : 190

Let $a_{1}, d_{1}$ be the first term and common difference of first A.P
Similarly $a_{2} a n d d_{2}$ be the first term and second difference of second A.P
Ratio of sum of n terms $= \frac{\frac{n}{2} (2 a_{1} + (n - 1) d_{1})}{\frac{n}{2} (2 a_{2} + (n - 1) d_{2})} = \frac{3 n + 8}{7 n + 15}$
$\Rightarrow \frac{(2 a_{1} + (n - 1) d_{1})}{(2 a_{2} + (n - 1) d_{2})} = \frac{3 n + 8}{7 n + 18}$
Substituting n = 25 gives
$\Rightarrow \frac{(2 a_{1} + (25 - 1) d_{1})}{(2 a_{2} + (25 - 1) d_{2})} = \frac{3 \times 25 + 8}{7 \times 25 + 15} \Rightarrow \frac{(2 a_{1} + 24 d_{1})}{(2 a_{2} + 24 d_{2})} = \frac{75 + 8}{175 + 15} \Rightarrow \frac{(a_{1} + 12 d_{1})}{(a_{2} + 12 d_{2})} = \frac{86}{190} \Rightarrow R a t i o o f 13^{t h} t e r m s = \frac{83}{190}$

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If the sum of first m terms of an AP is the same as the sum of its first n terms, show that the sum of its (m +n) terms is zero.

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

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