Asked by Anonymous

A g.p has a first term of "a", a common ratio of "r" and its 6th term is 768. Another G.P has a first term of "a", common ratio of 6r and its 3rd term is 3456. Evaluate "a and r".
Answered by mathhelper
1st GP
"6th term is 768" ---> ar^5 = 768

2nd GP, a = a, r = 6r
3rd term = 3456

a(6r)^2 = 3456
36a r^2 = 3456

36a r^2 / (ar^5) = 3456/768
36/r^3 = 9/2
9r^3 = 72
r^3 = 8
r = 2

back in ar^5 = 768
a(2^5) = 768
a = 24
Answered by Anonymous
Tn = a r^(n-1)
T6 = a r^5 = 768

T'n' = a (6r)^(n'-1)
T'3 = a (6r)^2 = 36 a r^2 = 3456

a r^5 /36 a r^2 = 768/3456
r^3 = 8
r = 2
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