The sum of three numbers which are consecutive terms an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.

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The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now from an A.P. Then the sum of the original three terms of the given G.P. is

The correct option is D 28Let terms bear,a,ar→G.P∴a3=512⇒a=88r+4,12,8r→A.P.24=8r+4+8rr=2,r=12r=2(4,8,16)r=12(16,8,4)Sum=28

The product of three consecutive terms of a $G . P .$ is $512 .$ If $4$ is added to each of the first and the second of these terms, the three terms now from an $A . P .$ Then the sum of the original three terms of the given $G . P .$ is

The correct option is D $28$
Let terms be
$\frac{a}{r}, a, a r \to G . P$
$∴ a^{3} = 512 \Rightarrow a = 8$
$\frac{8}{r} + 4, 12, 8 r \to A . P .$
$24 = \frac{8}{r} + 4 + 8 r$
$r = 2, r = \frac{1}{2}$
$r = 2 (4, 8, 16)$
$r = \frac{1}{2} (16, 8, 4)$
$S u m = 28$