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, ...
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
2 answers
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
d = 4
a = 8
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
(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
d = 4
a = 8
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
1 answer