The ratio of sum of n terms of two AP is 5n-3 divideb by 3n+23. Prove that their seventh terms are equal

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Solution

Let the two AP's have first terms as $a_{1}$ and $a_{2}$ and common difference as $d_{1}$ and $d_{2}$
7th terms are $\to a_{1} + 6 d_{1}$ and $a_{2} + 6 d_{2}$
Sum of 13 terms
$S_{13_{A}} = \frac{13}{2} {2 a_{1} + 12 d_{1}} = 13 (a_{1} + 6 d_{1}) S_{13_{B}} = \frac{13}{2} {2 a_{2} + 12 d_{2}} = 13 (a_{2} + 6 d_{2}) \frac{S_{13_{A}}}{S_{13_{B}}} = \frac{5 \times 13 - 3}{3 \times 13 + 23} (r a t i o g i v e n) = \frac{65 - 3}{39 + 23} = \frac{62}{62} = 1 ∴ \frac{13 (a_{1} + 6 d_{1})}{13 (a_{2} + 6 d_{2})} = 1 \Rightarrow a_{1} + 6 d_{1} = a_{2} + 6 d_{2}$ which is 7th term of both AP's.

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