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.
2 answers
Anyone to help solve this pliz
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
11 answers