Question

The sum of four numbers in GP is 60 and the a.m. of the first and the last is 18 the numbers are

Answer 1

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Answer 2

Answer:

Let the four terms of Gp be $a,ar,ar^2,ar^3$

The sum of four numbers in GP is 60

So,$a+ar+ar^2+ar^3=60$ ---A

We are given that  the a.m. of the first and the last is 18

So,$\frac{a+ar^3}{2}=18$

$a+ar^3=36$ --B

Substitute the value in A

$36+ar+ar^2=60$

$ar+ar^2=60-36=24$ ---C

Divide B and C

$\frac{a+ar^3}{ar+ar^2}=\frac{36}{24}$

$\frac{1+r^3}{r+r^2}=\frac{36}{24}$

$2r^3-3r^2-3r+2=0$

$r=-1,\frac{1}{2},2$

So, at r = 1/2

$a(1+\frac{1}{2}^3)=36$

$a=32$

$a,ar,ar^2,ar^3=32,16,8,4$

At r = 2

$a(1+2^3)=36$

$a=2$

$a,ar,ar^2,ar^3=32,64,128,256$