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
Jun 1, 2018
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
sn=a1−r
The question tells us that
16=a1−r→16(1−r)=a
Next we recall that the sum of the first n terms of a geometric progression is
sN=a(1−rn)1−r
15=a(1−r4)1−r
We can simplify the equation a little before combining it with the other one.
15=a(1−r2)(1+r2)1−r
15=a(1+r)(1−r)(1+r2)1−r
15(1+r)(r2+1)=a
We can now see that
16(1−r)=15(1+r)(r2+1)
16(1−r)(r3+r2+r+1)=15
16(r3+r2+r+1−r4−r3−r2−r)=15
16−16r4=15
1=16r4
116=r4
r=±12
We have two possible situations here.
**When **
16=a1−12→a=16(12)=8
The first four terms here are
8,4,2,1
When
16=a32→a=16(32)=24
The first four terms here are
24,−12,6,−3
Hopefully this helps!
