If the sum of the first n terms of two arithm | vedclass

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(C) For the first arithmetic progression $(AP_1)$: $3, 7, 11, 15, \ldots$Here,first term $a_1 = 3$ and common difference $d_1 = 7 - 3 = 4$.The sum of

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$55$

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(C) For the first arithmetic progression $(AP_1)$: $3, 7, 11, 15, \ldots$Here,first term $a_1 = 3$ and common difference $d_1 = 7 - 3 = 4$.The sum of $n$ terms is $S_{n_1} = \frac{n}{2} \{2a_1 + (n - 1)d_1\} = \frac{n}{2} \{2(3) + (n - 1)4\} = \frac{n}{2} \{6 + 4n - 4\} = \frac{n}{2} \{4n + 2\} = n(2n + 1)$.For the second arithmetic progression $(AP_2)$: $30, 33, 36, 39, \ldots$Here,first term $a_2 = 30$ and common difference $d_2 = 33 - 30 = 3$.The sum of $n$ terms is $S_{n_2} = \frac{n}{2} \{2a_2 + (n - 1)d_2\} = \frac{n}{2} \{2(30) + (n - 1)3\} = \frac{n}{2} \{60 + 3n - 3\} = \frac{n}{2} \{57 + 3n\} = \frac{3n(19 + n)}{2}$.Given $S_{n_1} = S_{n_2}$:$n(2n + 1) = \frac{3n(19 + n)}{2}$Since $n 
eq 0$,we divide by $n$:$2n + 1 = \frac{3(19 + n)}{2}$$2(2n + 1) = 3(19 + n)$$4n + 2 = 57 + 3n$$4n - 3n = 57 - 2$$n = 55$.

If the sum of the first $n$ terms of two arithmetic progressions $3, 7, 11, 15, \ldots$ and $30, 33, 36, 39, \ldots$ are equal,then find the value of $n$.

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