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

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(B) The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.Given the $7^{th}$ term,$a_7 = a + 6d = 40$.The sum of the first $n$ terms of an $

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$520$

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(B) The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.Given the $7^{th}$ term,$a_7 = a + 6d = 40$.The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.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 $S_{13} = 13 	imes 40 = 520$.

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

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