Question

in an arithmetic sequence the common difference is equal to 2.the first term is also the first term of a geometric sequence. the sum of the first 3 terms of an arithmetic sequence and the sum of the first 9 terms of an arithmetic sequence form the 2nd and 3rd terms of a geometric sequence respectively determine the first 3 terms of the geometric sequence...How do I go about solving this?

Answer 1

first term: a
common difference = 2

sum of first 3 terms
= a + a+d + a+2d
= 3a + 6
sum of first 9 terms
= (9/2)(2a + 8(2))
= 9(a + 8)
= 9a + 72

so 3a+6 = ar
and 9a+72 = ar^2

square 3a+6 = ar^2
9a^2 + 36a + 36 = a^2r^2
r^2= (9a^2 + 36a + 36)/a^2

from the other equation,
r^2 = (9a+72)/a

(9a^2 + 36a + 36)/a^2 = (9a+72)/a
both sides times a
(9a^2 + 36a + 36)/a = (9a+72)
cross-multiply
9a^2 + 36a + 36 = 9a^2 + 72a
36a = 36
a = 1

in 3a+6 = ar
3+6 = r
r = 9

So the first 3 terms of the GS are 1 , 9, and 81

check:
sum of first 3 terms of AS=1 + 3 + 5
sum of first 9 terms of AS = (9/2)(2 + 16) = 81
Well, that checked out nicely.