Asked by Priscilla
The 3rd and 4th term of geometric progression are 4 and 8 respectively find:
The first term and common ratio,
The sum of the first 10 term,
The sum of infinity.
The first term and common ratio,
The sum of the first 10 term,
The sum of infinity.
Answers
Answered by
Priscilla
Anyone to help solve this pliz
Answered by
mathhelper
Just apply the definitions:
3rd term = ar^2 = 4
4th term = ar^3 = 8
divide them
ar^3/(ar^2) = 8/4
r = 1/2
back in ar^2 = 4
a(1/4) = 4
a = 16
sum(10) = a(1 - r^10)/(1-r)
= 16( 1 - 1/1024)/(1 - 1/2)
= 16(1023/1024)(2/1) = 1023/32
sum(all terms) = a/(1-r)
= 16/(1/2) = 32
3rd term = ar^2 = 4
4th term = ar^3 = 8
divide them
ar^3/(ar^2) = 8/4
r = 1/2
back in ar^2 = 4
a(1/4) = 4
a = 16
sum(10) = a(1 - r^10)/(1-r)
= 16( 1 - 1/1024)/(1 - 1/2)
= 16(1023/1024)(2/1) = 1023/32
sum(all terms) = a/(1-r)
= 16/(1/2) = 32
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