Question

find three numbers in Gp whose sum is 52 and the sum of whose product in pairs is 624

Answer 1

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Answer 2

Answer:  The required numbers are 4, 12 and 36.

Step-by-step explanation:  We are given to find the three numbers in G.P. whose sum is 52 and the sum of whose product in pairs is 624.

Let a, ar, ar² be the given three numbers in G.P., with first term a and common ratio r.

Then, according to the given information, we have

$a+ar+ar^2=52\\\\\Rightarrow a(1+r+r^2)=52~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$a\times ar+ar\times ar^2+ar^2\times a =624\\\\\Rightarrow a^2r+a^2r^3+a^2r^2=624\\\\\Rightarrow a^2r(1+r+r^2)=624~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Dividing equation (ii) by equation (i), we get

$\dfrac{a^2r(1+r+r^2)}{a(1+r+r^2)}=\dfrac{624}{52}\\\\\\\Rightarrow ar=12\\\\\Rightarrow r=\dfrac{12}{a}~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Substituting the value of r from equation (iii) in equation (i), we get

$a\left(1+\dfrac{12}{a}+\dfrac{12^2}{a^2}\right)=52\\\\\\\Rightarrow a+12+\dfrac{144}{a}=52\\\\\Rightarrow a+\dfrac{144}{a}=40\\\\\Rightarrow a^2+144=40a\\\\\Rightarrow a^2-40a+144=0\\\\\Rightarrow a^2-36a-4a+144=0\\\\\Rightarrow a(a-36)-4(a-36)=0\\\\\Rightarrow (a-4)(a-36)=0\\\\\Rightarrow a-4=0,~~~a-36=0\\\\\Rightarrow a=4,~36.$

If a = 4, then from equation (iii), we get

$r=\dfrac{12}{4}=3.$

And, if a = 36, then from equation (iii), we get

$r=\dfrac{12}{36}=\dfrac{1}{3}.$

Therefore, the required numbers are

$a,~ar,~ar^2=4,~4\times3,~4\times3^2=4,~12,~36,\\or\\\\a~ar,~ar^2=36,~36\times\dfrac{1}{3},~36\times\dfrac{1}{3^2}=36,~12,~4.$

Thus, the required numbers are 4, 12 and 36.