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Tuesday, September 16, 2008

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

A) 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 r>1 (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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1 comment:

Unknown said...

thanks a lot!!!!

phew !!! homework done with that !!!