Question

What is the sum of the first 10 terms of an A.P. 41,36,31,36​

Answer 1

Answer:

$AP:-41, 36, 31,36 \\ a=41, d=36-41=(-5),n=10 \\ S_{n}= \frac{n}{2} [2a+(n-1)d] \\ S_{10}= \frac{10}{2} [2(41)+(10-1)( - 5)] \\ S_{10}=5[82+(9)( - 5)] \\ S_{10}=5[82+( - 45)] \\ S_{10}=5(82 - 45) \\ S_{10}=5(37) \\ \boxed{ \huge{ }S_{10}=185}\\ Therefor, \: the \: sum \: of \: 10 \: terms \: of \: the \: AP \: is \: 185.$