If the first term of an A.P. is 3 and the sum | vedclass

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(D) Let the first term be $a = 3$ and the common difference be $d$.The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-

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$\frac{1}{6}$

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(D) Let the first term be $a = 3$ and the common difference be $d$.The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.The sum of the first $25$ terms is $S_{25} = \frac{25}{2}[2(3) + 24d] = \frac{25}{2}[6 + 24d] = 25(3 + 12d) = 75 + 300d$.The sum of the next $15$ terms is $S_{26-40} = S_{40} - S_{25}$.$S_{40} = \frac{40}{2}[2(3) + 39d] = 20[6 + 39d] = 120 + 780d$.Given $S_{25} = S_{40} - S_{25}$,which implies $2S_{25} = S_{40}$.$2(75 + 300d) = 120 + 780d$.$150 + 600d = 120 + 780d$.$150 - 120 = 780d - 600d$.$30 = 180d$.$d = \frac{30}{180} = \frac{1}{6}$.

If the first term of an $A.P.$ is $3$ and the sum of its first $25$ terms is equal to the sum of its next $15$ terms,then the common difference of this $A.P.$ is:

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