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 $d$.Given that the sum of the first $25$ terms is equal to the sum of the next $15$ ter

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$\frac{1}{6}$

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(D) Let the first term be $a = 3$ and the common difference be $d$.Given that the sum of the first $25$ terms is equal to the sum of the next $15$ terms.Let $S_n$ denote the sum of the first $n$ terms.The sum of the first $25$ terms is $S_{25}$.The sum of the next $15$ terms is $S_{40} - S_{25}$.According to the problem,$S_{25} = S_{40} - S_{25}$,which implies $2S_{25} = S_{40}$.Using the formula $S_n = \frac{n}{2}[2a + (n-1)d]$:$2 	imes \frac{25}{2}[2(3) + (25-1)d] = \frac{40}{2}[2(3) + (40-1)d]$$25[6 + 24d] = 20[6 + 39d]$Divide both sides by $5$:$5[6 + 24d] = 4[6 + 39d]$$30 + 120d = 24 + 156d$$30 - 24 = 156d - 120d$$6 = 36d$$d = \frac{6}{36} = \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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