Question

The sum of three numbers in g.p is 56.If we subtract 1,7,21 from these numbers in that order,we obtain an a.p.Find the numbers.

Answer 1

Let the three numbers in G.P. be a,ar,ar²

It given that a+ar+ar²=56 ....(1)

It is also given that (a−1), (ar−7) and (ar²−21) are in A.P.

We know that if any three numbers x,y,z are in A.P., then 2y=x+z.

⇒2(ar−7)=(a−1)+(ar²−21)

⇒2ar−14=a+ar²−22

From (i) we get 
(a+ar²=56−ar)

∴2ar=(56−ar)−22+14

⇒ 3ar=48
⇒ ar = 16
and a = 16/r

Substituting the value of ar and a in (i) we get

a+16+ar²=56

⇒a(1+r²)=40

⇒(16/r)(1+r²)=40

2+2r²=5r

⇒2r²−5r+2=0

⇒2r²−4r−r+2=0

⇒2r(r−2)−1(r−2)=0

⇒r=2 or r=1/2

Substituting the value of r=2, we get a = 16/2 = 8 

and if r=1/2 we get a = 16/(1/2) = 32

∴ The numbers are 8 ,16 and 32