Question

the sum of the first nine terms of an arithmetic sequence 216. The 1st, 3rd and 7th terms form the first three terms of a geometric sequence therefore find the common difference

Answer 1

9/2 (2a+8d) = 216
(a+2d)/a = (1+6d)/(a+2d)

One choice is

24, 24, 24, ...

Probably not what you want. So, let's try

8, 12, 16, ...

Answer 2

sum(9) = (9/2)(2a + (n-1)d)
(9/2)(2a + 8d) = 216
9(a + 4d) = 216
a + 4d = 24

term1 = a
term3 = a+2d
term7 = a + 6d

they form a GP, so
(a+2d)/a = (a+6d)/(a+2d)
a^2 + 6ad = a^2 + 4d + 4d^2
2ad = 4d^2
a = 2d , assuming d ≠ 0

back into a+4d = 24
2d + 4d = 24
<b>d = 4
a = 8</b>

check:
terms are
8 , 12, 16, 20 , 24, 28, 32...

term7/term3 = term3/term1
32/16 = 16/8
true!

sum(9) = (9/2)(16 + 8(4)) = 216 , ok