The sum of three numbers in $G . P .$ is $14$ . If the first two terms are each increased by $1$ and the third term decreased by $1$ , the resulting numbers are in $A . P .$ Find the numbers.

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Solution

$a (1 + r + r^{2}) = 14$
and $2 (a r + 1) = (a + 1) + (a r^{2} - 1)$
or $a (r^{2} - 2 r + 1) = 2$
Divide $(1)$ and $(2)$ and simplifying,
$2 r^{2} - 5 r + 2 = 0$ $∴$ $r = 2, \frac{1}{2}$ and hence $a = 2, 8$ .
$∴$ Numbers are $2, 4, 8,$ or $8, 4, 2$ .

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The sum of three numbers in G.P. is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in A.P. Find the numbers.

a(1+r+r2)=14and 2(ar+1)=(a+1)+(ar2−1)or a(r2−2r+1)=2Divide (1) and (2) and simplifying,2r2−5r+2=0 ∴ r=2,12 and hence a=2,8.∴ Numbers are 2,4,8, or 8,4,2.

The sum of three numbers in $G . P .$ is $14$ . If the first two terms are each increased by $1$ and the third term decreased by $1$ , the resulting numbers are in $A . P .$ Find the numbers.

$a (1 + r + r^{2}) = 14$
and $2 (a r + 1) = (a + 1) + (a r^{2} - 1)$
or $a (r^{2} - 2 r + 1) = 2$
Divide $(1)$ and $(2)$ and simplifying,
$2 r^{2} - 5 r + 2 = 0$ $∴$ $r = 2, \frac{1}{2}$ and hence $a = 2, 8$ .
$∴$ Numbers are $2, 4, 8,$ or $8, 4, 2$ .