Question

Find S18 for geometric series given a5=-6 and a2= -48

Answer 1

an = a1r^(n-1)

a5 = a1 r^(5-1)

-6 =a1 r^4

a2 = a1 r^(2-1)

-48 = a1 r

divide

-6 =a1 r^4

----------------    yields   1/8 = r^3      take the cube root  or each side

-48 = a1 r                     1/2 = r

an = a1r^(n-1)

an = a1 (1/2)^ (n-1)

-48 = a1 (1/2) ^1

divide by 1/2

-96 = a1

an = -96 (1/2)^ (n-1)

the sum

Sn = a1[(r^n - 1/(r - 1)]

S18 = -96 [( (1/2) ^17 -1/ (1/2 -1)]

       =-96 [ (1/2) ^ 17 -1 /-1/2]

      = 192 * [-131071/131072]

 approximately -192