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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## Tuesday, September 16, 2008

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## 1 comment:

thanks a lot!!!!

phew !!! homework done with that !!!

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