Given an a.p where a= 5
n = 16 and d= 1
We are asked sum of A.P
which is
S= \(\frac{n}{2}(2a+(n-1)d)\)
= \(\frac{16}{2}(10+15*1)\)
= 200
hence answer B
n = 16 and d= 1
We are asked sum of A.P
which is
S= \(\frac{n}{2}(2a+(n-1)d)\)
= \(\frac{16}{2}(10+15*1)\)
= 200
hence answer B







