Question

Determine the sum of first 35 terms of an A.P. If t2=1 and t7=22.

Answer 1

Given:

• In an A.P. second term is 1 and seventh term is 22.

To find:

• Sum of the first 35 terms.

Solution:

$\boxed{\rm{t_{n}=a+(n-1)d}} \\ \\ \sf{According \ to \ the \ first \ condition} \\ \\ \sf{a+d=1...(1)} \\ \\ \sf{According \ to \ the \ second \ condition} \\ \\ \sf{a+6d=35...(2)} \\ \\ \sf{Subtract \ eq(1) \ from \ eq(2), \ we \ get} \\ \\ \sf{5d=34} \\ \\ \sf{\therefore{d=\dfrac{34}{5}=6.8}} \\ \\ \sf{Substitute \ d=6.8 \ in \ equation (1), \ we \ get} \\ \\ \sf{a+6.8=1} \\ \\ \sf{\therefore{a=-5.8}} \\ \\ \boxed{\sf{S_{n}=\dfrac{n}{2}[2a+(n-1)d]}} \\ \\ \sf{\therefore{S_{35}=\dfrac{35}{2}[2(-5.8)+(35-1)6.8]}} \\ \\ \sf{\therefore{S_{35}=17.5[-11.6+231.2]}} \\ \\ \sf{\therefore{S_{35}=17.5(219.6)}} \\ \\ \sf{\therefore{S_{35}=3843}} \\ \\ \sf{\therefore{Sum \ of \ 35 \ terms \ in \ the \ A.P. \ is \ 3843.}}$