JEE MAIN - Mathematics (2019 - 11th January Morning Slot - No. 21)
Let a1, a2, . . . . . ., a10 be a G.P. If $${{{a_3}} \over {{a_1}}} = 25,$$ then $${{{a_9}} \over {{a_5}}}$$ equals
53
2(52)
4(52)
54
Explanation
a1, a2, . . . . ., a10 are in G.P.,
Let the common ratio be r
$${{{a_3}} \over {{a_1}}} = 25 \Rightarrow {{{a_1}{r^2}} \over {{a_1}}} = 25 \Rightarrow {r^2} = 25$$
$${{{a_9}} \over {{a_5}}} = {{{a_1}{r^8}} \over {{a_1}{r^4}}} = {r^4} = {5^4}$$
Let the common ratio be r
$${{{a_3}} \over {{a_1}}} = 25 \Rightarrow {{{a_1}{r^2}} \over {{a_1}}} = 25 \Rightarrow {r^2} = 25$$
$${{{a_9}} \over {{a_5}}} = {{{a_1}{r^8}} \over {{a_1}{r^4}}} = {r^4} = {5^4}$$
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