The sum of the first n terms of an AP whose f | vedclass

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(B) Given that,for the first $AP$,the first term $a = 8$ and the common difference $d = 20$.Let the number of terms be $n$.The sum of the first $n$ te

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$11$

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(B) Given that,for the first $AP$,the first term $a = 8$ and the common difference $d = 20$.Let the number of terms be $n$.The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$.$S_n = \frac{n}{2}[2(8) + (n-1)20] = \frac{n}{2}[16 + 20n - 20] = \frac{n}{2}[20n - 4] = n(10n - 2) = 10n^2 - 2n$ .....$(i)$For the second $AP$,the first term $a' = -30$ and the common difference $d' = 8$.The sum of the first $2n$ terms is $S'_{2n} = \frac{2n}{2}[2a' + (2n-1)d']$.$S'_{2n} = n[2(-30) + (2n-1)8] = n[-60 + 16n - 8] = n[16n - 68] = 16n^2 - 68n$ .....$(ii)$According to the problem,$S_n = S'_{2n}$.$10n^2 - 2n = 16n^2 - 68n$$6n^2 - 66n = 0$$6n(n - 11) = 0$Since $n$ must be a positive integer,$n = 11$.

The sum of the first $n$ terms of an $AP$ whose first term is $8$ and the common difference is $20$ is equal to the sum of first $2n$ terms of another $AP$ whose first term is $-30$ and the common difference is $8$. Find $n$.

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