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Nth Term of GP

Let a1, a2, …...

Question

Let $a_{1}, a_{2}, \dots, a_{10}$ be a G.P. If $\frac{a_{3}}{a_{1}} = 25$ , then $\frac{a_{9}}{a_{5}}$ equals:

A

2 (5)^{2}

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B

4 (5^{2})

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C

5^{3}

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D

5^{4}

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Solution

The correct option is D $5^{4}$
$\frac{a_{3}}{a_{1}} = 25 \Rightarrow \frac{a r^{2}}{a} = 25$
$\Rightarrow r^{2} = 25$
$∴ \frac{a_{9}}{a_{5}} = \frac{a r^{8}}{a r^{4}} = r^{4} = 25^{2} = 5^{4}$

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Similar questions

a_{1}, a_{2}, . . . . ., a_{10}

\frac{a_{3}}{a_{1}} = 25

\frac{a_{9}}{a_{5}}

a_{1}, a_{2}, \dots, a_{10}

\frac{a_{3}}{a_{1}} = 25

\frac{a_{9}}{a_{5}}

a, a_{1}, a_{2} . . . . . . . . . . . . . a_{10}, b

are in A.P. and

a, g_{1}, g_{2} . . . . . . g_{10}, b

are in G.P. and

h

is the H.M between

a

b

\frac{a_{1} + a_{2} + . . . . . . . . . + a_{10}}{g_{1} g_{10}} + \frac{a_{2} + a_{3} + . . . . . . . . . + a_{9}}{g_{2} g_{9}} + . . . . . . . . . . . + \frac{a_{5} + a_{6}}{g_{5} g_{6}}

a_{1}, a_{2}, a_{3}, . . . .

are in A.P. such that

a_{1}, a_{5}, a_{10} + a_{15} + a_{20} + a_{24} = 225

a_{1} + a_{2} + a_{3} + . . . + a_{24}

a_{1}, a_{2}, a_{3}, . . . . . . . . i s a n A . P . s u c h t h a t a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225

a_{1} + a_{2} + a_{3} + . . . . . . . + a_{23} + a_{24} i s e q u a l t o

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