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InterviewSolution
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The sum of n terms of two arithmetic progression is in the ratio 2n+1 : 2n-1. Find the ratio of `10^th` term. |
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Answer» Let first term in first AP is `a_1` and common difference is `d_1`. Let first term in second AP is `a_2` and common difference is `d_2`. Then, we are given, `S_1/S_2 = (2n+1)/(2n-1)` `(n/2(2a_1+(n-1)d_1))/(n/2(2a_2+(n-1)d_2)) = (2n+1)/(2n-1)` `=>(2a_1+(n-1)d_1)/(2a_2+(n-1)d_2) = (2n+1)/(2n-1)->(1)` Now, we have to find ratio of `10th ` terms of both AP. If we put `n = 19` in (1), `(2a_1+18d_1)/(2a_2+18d_2) = 39/37` `=> (a_1+9d_1)/(a_2+9d_2) = 39/37->(2)` `10th` term of an AP is given by, `T_10 = a+9d` So, equation `(2)` gives us the required ratio that is `39:37`. |
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