# The sum to infinity if a GP is 16 and the sum of the first 4 terms is 15. Find the first four terms?

##### 1 Answer

The first four terms may either be

#8, 4, 2, 1#

OR

#24, -12, 6, -3#

#### Explanation:

We know that the sum of an infinite geometric series is

#s_n = a/(1 - r)#

The question tells us that

#16 = a/(1 - r) -> 16(1 - r) = a#

Next we recall that the sum of the first n terms of a geometric progression is

#s_N = (a(1 - r^n))/(1 -r)#

#15 = (a(1 - r^4))/(1 - r)#

We can simplify the equation a little before combining it with the other one.

#15 =(a(1 - r^2)(1 + r^2))/(1 -r)#

#15 = (a(1 + r)(1 - r)(1 + r^2))/(1 - r)#

#15/((1 + r)(r^2 + 1)) = a#

We can now see that

#16(1 - r) = 15/((1 +r)(r^2 + 1))#

#16(1 -r)(r^3 + r^2+ r + 1) = 15#

#16(r^3 + r^2 + r + 1 - r^4 - r^3 - r^2 - r) = 15#

#16 - 16r^4 = 15#

#1 = 16r^4#

#1/16 = r^4#

#r = +- 1/2#

We have two possible situations here.

**When **

#16 = a/(1 - 1/2) -> a = 16(1/2) = 8#

The first four terms here are

#8, 4, 2, 1#

**When #r =-1/2#**

#16 = a/(3/2) -> a =16(3/2) = 24#

The first four terms here are

#24, -12, 6, -3#

Hopefully this helps!