What is the first term of the Geometric series with a sum of 1022 the last term is 512 and the common ratio is 2?

Answer 1

The sum of the first n terms of a geometric series is

Sn = [t1(1 - r^n)]/(1 - r) (1)

where r is the common ratio and t1 is the first term.

The last term is tn = (t1)[r^(n - 1)] (2).

We know the sum and the common ratio, but we don't know how many terms are in this series. So, in order to find the first term t1 we need to find n.

From formula (1):

Sn = [t1(1 - r^n)]/(1 - r)

Sn(1 - r) = t1(1 - r^n)

t1 = [Sn(1 - r)]/(1 - r^n)

From formula (2):

tn = (t1)[r^(n - 1)]

t1 = tn/[r^(n - 1)] (3)

So we have:

[Sn(1 - r)]/(1 - r^n) = tn/[r^(n - 1)] substitute Sn = 1.022, r = 2, and tn = 512;

[1,022(1 - 2)]/(1 - 2^n) = 512/[2^(n - 1)]

-1,022/(1 - 2^n) = 512/[(2^n/2^1)]

-1,022/(1 - 2^n) = 512(2/2^n) divide by 2 to both sides;

-511/(1 - 2^n) = 512/2^n cross multiply;

(-511)(2^n) = (512)(1 - 2^n)

(-511)(2^n) = 512 - (512)(2^n) add (512)(2^n) to both sides;

2^n = 512 rewrite this in its logarithmic form;

n = log(base 2) of 512

n = log 512/log 2

n = 9

From formula (3):

t1 = tn/[r^(n - 1)] substitute tn = 512, r = 2, and n = 9;

t1 = 512/[2^(9 - 1)]

t1 = 512/2^8

t1 = 512/256

t1 = 2

Thus the first term is 2.

Check:

1,022 = [2(1 - 2^9)]/(1 - 2) ?

1,022 = [2(1 - 512)]/(-1) ?

1,022 = ((2)(-511)]/(-1) ?

1,022 = -1,022/-1 ?

1,022 = 1,022 True