A) 520
B) 53
C) 2080
D) 1040
Correct Answer: A
Solution :
In A.P. \[{{T}_{n}}=a+(n-1)\,d\] \[{{T}_{7}}=a+6d=40\] [Given] and \[{{S}_{n}}=\frac{n}{2}\,[2a+(n-1)\,d]\] \[\therefore \] \[{{S}_{13}}=\frac{13}{2}\,[2a+12d]\] \[=13\,[a+6d]\] \[=13\times 40\] \[\because \,\,[a+6d=40]\] = 520You need to login to perform this action.
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