A) \[10\times 41\]
B) \[\frac{10\times 41}{2}\]
C) \[10\times 81\]
D) \[41\times 81\]
Correct Answer: C
Solution :
[c] \[n=81\] \[{{t}_{41}}=a+(n-1).d\] \[10=a+40d\] \[{{S}_{81}}=\frac{n}{2}\left\{ 2a+\left( n-1 \right).d \right\}=n\left\{ 2a+80d \right\}\] \[=\frac{81}{2}\times \left\{ a+40d \right\}=81\times 10.\] Hence, option [c] is correct.
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