The sum of three numbers in G.P. is 56. If we | vedclass

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(A) Let the three numbers in $G.P.$ be $a, ar, ar^2$.From the given condition,$a + ar + ar^2 = 56 \Rightarrow a(1 + r + r^2) = 56$ ........$(1)$Subtra

Vedclass · Accepted Answer

$8, 16, 32$

Vedclass · Answer

(A) Let the three numbers in $G.P.$ be $a, ar, ar^2$.From the given condition,$a + ar + ar^2 = 56 \Rightarrow a(1 + r + r^2) = 56$ ........$(1)$Subtracting $1, 7, 21$ from these numbers,we get $a-1, ar-7, ar^2-21$,which are in $A.P.$Therefore,$(ar-7) - (a-1) = (ar^2-21) - (ar-7)$$ar - a - 6 = ar^2 - ar - 14$$ar^2 - 2ar + a = 8$ $\Rightarrow a(r^2 - 2r + 1) = 8$ $\Rightarrow a(r-1)^2 = 8$ ........$(2)$Dividing $(1)$ by $(2)$:$\frac{a(1+r+r^2)}{a(r-1)^2} = \frac{56}{8} = 7$$1 + r + r^2 = 7(r^2 - 2r + 1)$$1 + r + r^2 = 7r^2 - 14r + 7$$6r^2 - 15r + 6 = 0 \Rightarrow 2r^2 - 5r + 2 = 0$$(2r-1)(r-2) = 0$If $r=2$,$a(2-1)^2 = 8 \Rightarrow a=8$. The numbers are $8, 16, 32$.If $r=1/2$,$a(1/2-1)^2 = 8$ $\Rightarrow a(1/4) = 8$ $\Rightarrow a=32$. The numbers are $32, 16, 8$.

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 arithmetic progression. Find the numbers.

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