A1,a2,a3,...,a40 are the terms of am AP and a1+a5+a15+a26+a36+a40=105 then find a1+a2+a3+....+a40 ?

Question

4209 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-01T18:05:03

# 700#.

Suppose that the common difference of the AP in question is #d#.

#:. a_n=a_1+(n-1)d, n in NN#.

Given that, #a_1+a_5+a_15+a_26+a_36+a_40=105,#

#rArr a_1+{a_1+4d}+{a_1+14d}+{a_1+25d}+{a_1+35d}+{a_1+39d}=105#.

#:. 6a_1+117d=105#.

Dividing by #3, 2a_1+39d=35................(star)#.

Recall that, # a_1+a_2+...+a_n=n/2{2a_1+(n-1)d}#.

#:. a_1+a_2+...+a_40=40/2{2a_1+39d}#,

#=20(35)...............[because, (star)]#,

#=700#.