Question

How many terms of the arithmetic sequence
99, 97, 95, ... must be added to get 900?​

Answer 1

Here $Sn=900 \\ a=99 \\ d=-2 \\ 900=n*{198+(n-1)*(-2)}/2 \\ 1800=198n+(n^2-n)-2900=99n-n^2+n \\ n^2–100n+900=0 \\ (n-10)(n-90)=0 \\ So \: n=10,90$

It indicates that sum of 11th term to 90th is zero.

We must add either first 10 terms or first 90 terms to get 900

Answer 2

Answer:

thus sum of Frist 10 terms and of 90 terms =900