If the 7^{th} term of an A.P. is 40,then the | vedclass

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(B) Given that the $7^{th}$ term of an $A.P.$ is $40$. $a + (7 - 1)d = 40 \implies a + 6d = 40$. We need to find the sum of the first $13$ terms,denot

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$520$

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(B) Given that the $7^{th}$ term of an $A.P.$ is $40$. $a + (7 - 1)d = 40 \implies a + 6d = 40$. We need to find the sum of the first $13$ terms,denoted by $S_{13}$. The formula for the sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n - 1)d]$. For $n = 13$,$S_{13} = \frac{13}{2}[2a + (13 - 1)d] = \frac{13}{2}[2a + 12d]$. Factoring out $2$,we get $S_{13} = \frac{13}{2} 	imes 2(a + 6d) = 13(a + 6d)$. Substituting the value $a + 6d = 40$,we get $S_{13} = 13 	imes 40 = 520$.

If the $7^{th}$ term of an $A.P.$ is $40$,then the sum of the first $13$ terms is:

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