How do you find the sum of the geometric series $3-6+12-...$ to 7 terms?

Question

5698 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-19T10:41:34

$"Reqd. Sum" =129.$

In the given G.S.,

we have, $a="the "1^(st)" term"=3,"& the common ratio "r=-6/3=-2$

Hence, the sum $S_n$ of its first $n$ terms is given by,

$S_n=(a(1-r^n))/(1-r)$

$:." The Reqd. Sum S_7"={3(1-(-2)^7)}/{1-(-2)}={3(1+128)}/3=129$ .

Enjoy Maths.!

How do you find the sum of the geometric series 3-6+12-...3-6+12-... to 7 terms?