If the sum of the first 2n terms of the arith | vedclass

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(C) For the first arithmetic progression $2, 5, 8, \dots$,the first term $a_1 = 2$ and common difference $d_1 = 3$. The sum of the first $2n$ terms is

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$11$

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(C) For the first arithmetic progression $2, 5, 8, \dots$,the first term $a_1 = 2$ and common difference $d_1 = 3$. The sum of the first $2n$ terms is given by:$S_{2n} = \frac{2n}{2} [2(2) + (2n - 1)3] = n(4 + 6n - 3) = n(6n + 1) \dots (1)$For the second arithmetic progression $57, 59, 61, \dots$,the first term $a_2 = 57$ and common difference $d_2 = 2$. The sum of the first $n$ terms is given by:$S'_n = \frac{n}{2} [2(57) + (n - 1)2] = \frac{n}{2} [114 + 2n - 2] = \frac{n}{2} [2n + 112] = n(n + 56) \dots (2)$Given that $S_{2n} = S'_n$,we have:$n(6n + 1) = n(n + 56)$Since $n 
eq 0$,we can divide by $n$:$6n + 1 = n + 56$$5n = 55$$n = 11$

If the sum of the first $2n$ terms of the arithmetic progression $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of the arithmetic progression $57, 59, 61, \dots$,then $n = \dots$

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