A number a, its square and its cube are in GP. What will be the third term of AP, for which first tw

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13 Dec 2011 AP and GP

A number a, its square and its cube are in GP. What will be the third term of AP, for which first two terms are obtained by taking the difference of terms of GP in 2nd-1st and 3rd-2nd manner?

OPtion

1) 6*sum of all numbers upto (a-1)
2) 6*sum of squares of all numbers upto (a-1)
3) a+a^2+a^3
4) 1+a+a^2+a^3
5) 6*sum of all numbers upto a
6) 6*sum of squares of all numbers upto a
7) 6*sum of cubes of all numbers upto (a-1)
8) 6*sum of cubes of all numbers upto a
9) always a fixed number
10) none of these

Solution

Let the 3 numbers are a, a^2 and a^3
so differences = a^2-a and a^3-a^2
third term of the AP, for which first 2 terms are a^2-a and a^3-a^2 will be 2(a^3-a^2)-(a^2-a) = 6*[a(a-1)(2a-1)/6]
= 6*sum of squares of all numbers upto (a-1)

Correct Option: 2 Best Solution (7)

shloka   12 years AGO

1st & 2nd terms of AP are a^2-a and a^3-a^2
3rd term of AP = a^3-a^2 + (a^3-a^2 - (a^2-a))
Solving we get= a(2a-1)(a-1) --- Equation1
Sum of the Squares of all n Numbers =(n+1)(2n+1)/6
Applying above formula in Equation1 we get
6*((a-1)*a*(2a-1))/6 = 6*sum of squares of all numbers upto (a-1) [Option2]

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Datla Kanthi   12 years AGO

AP is a, a^2,a^3,...
GP is a^2-a,a^3-a,...

so third term 2a^3-a^2+a

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Shubham   12 years AGO

a^3-a^2 +(a^3-a^2-a^2+a)

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Akash M T   12 years AGO

by trial and error method

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Manish Kumar   12 years AGO

a, a2, a3 are in GP

a2-a, a3-a2 are two term of AP

3rd term=a(a-1)(2a-1)

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sandeep potnuru   12 years AGO

the third term is 2a^3-3a^2+a
so option 2 is correct since,sum of squares of all numbers upto (a-1)=2a^3-3a^2+a/6.

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sridhar reddy   12 years AGO

by given conditions a^2,a^3, a^4are in GP.
(a^3-a^2),(a^4-a^3),x are in AP
then 2(a^4-a^3)=(a^3-a^2)+x
x = (a-1)*a*(2a-1)
= (a-1)*((a-1)+1)*(2(a-1)+1)
x = 6*sum of squares of all numbers upto (a-1)

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