JEE MAIN - Mathematics (2019 - 12th January Morning Slot - No. 22)
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 form an A.P. Then the sum of the original three terms of the given G.P. is :
36
28
32
24
Explanation
Let terms are $${a \over r},a,ar \to G.P$$
$$ \therefore $$ $${a^3}$$ = 512 $$ \Rightarrow $$ a = 8
$${8 \over r} + 4,12,8r \to A.P.$$
24 = $${8 \over r} + 4 + 8r$$
r = 2, r = $${1 \over 2}$$
r = 2(4, 8, 16)
r = $${1 \over 2}$$ (16, 8, 4)
Sum = 28
$$ \therefore $$ $${a^3}$$ = 512 $$ \Rightarrow $$ a = 8
$${8 \over r} + 4,12,8r \to A.P.$$
24 = $${8 \over r} + 4 + 8r$$
r = 2, r = $${1 \over 2}$$
r = 2(4, 8, 16)
r = $${1 \over 2}$$ (16, 8, 4)
Sum = 28
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