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

1 Answer
Aug 1, 2018

700700.

Explanation:

Suppose that the common difference of the AP in question is dd.

:. 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.