A) 53
B) 520
C) 1040
D) 2080
Correct Answer: B
Solution :
\[{{7}^{th}}\]term of an A.P. \[=40\] \[a+6d=40\] \[{{S}_{13}}=\frac{13}{2}[2a+(13-1)d]=\frac{13}{2}[2(a+6d)]\] \[=\frac{13}{2}\ .\ 2\ .\ 40=520\].
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