Question

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

Answer 1

`"Reqd. Sum" =129.`

Explanation:

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.!