In an increasing G.P the sum of the first and last term is 66,the product of the second and the last but one is 128 and the sum of the terms is 126.How many terms are there in the progression?

Let the GP be : a, ar, ar^2 ....a.r^n

a + a.r^n = 66 = a(1 + r^n)
ar.a.r^(n-1) = 128 = a^2.r^n
So 66 = a (1 + 128/a^2)
66a = a^2 + 128
a^2 - 66a + 128 = 0
a = 66 +- rt(66^2 - 4(128)) / 2
a = 66 +- 62 / 2
a = 2, 64
Since, 66 = a(1 + r^n)
66 = 2(1 + r^n) OR 66 = 64 (1 + r^n)
Since r1 (increasing GP) so we take
66 = 2(1 + r^n)
r^n = 32

And,

a(r^(n+1) - 1) / (r - 1) = 126
2(32r - 1) / (r-1) = 126
63r - 63 = 32r - 1
31r = 62
r = 2
and r^n = 32
2^n = 32
so
n = 5.
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The number of terms is n+1=6.
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Please explain what is the meaning of this part of the question , " the product of the second and the last but one term "?
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