Question

The sum of the first three terms of geometric sequence is 14. If the first term is 2, find the possible values of the sum of the first 5 terms.

Answer 1

Geometric series are in the form of

$a +a*r +a*r^2+...$

Where a is the first term and r is the common ratio .

And it is given that

$2 +2*r +2*r^2=14$

$2r^2+2r-12=0$

$r^2 +r-6=0$

$(r+3)(r-2)=0$

r=-3,2

So the first five terms are

$2+2(-3)+2(-3)^2+2(-3)^3+2(-3)^4 or 2+2(2)+2(2)^2+2(2)^3+2(2)^4$

= 2-6+18-54+162 or 2+4+8+16+32

= 122 or 62

Answer 2

We have a first term of 2 and a common ratio of r. We're told

$2 + 2r + 2r^2 = 14$

$r^2 +r - 6 = 0$

$(r+3)(r-2)=0$

$r=-3 \textrm{ or } r=2$

If r=-3 the fourth term is 2((-3)³) and the fifth 2((-3)^4) for a sum of

14 - 54 + 162 = 122

If r=2 the remaining terms are 2(2^3)=16 and 2(2^4)=32 for a sum of

14 + 16 + 32 = 62

Answer: 62 or 122