Let a1- a2- - a10 be a G-P- If a3-a1-25- then a9-a5 equals-A- 252B- 452C- 53D- 54

The correct option is D 5^4a_3a_1=25 ⇒ar^2a=25 ⇒ r^2=25 ∴a_9a_5=ar^8ar^4=r^4=25^2=5^4 ...

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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:

2 (5)^{2}

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4 (5^{2})

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5^{3}

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5^{4}

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Solution

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Similar questions

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

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

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

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

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

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

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

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

h

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}, . . . . . . . . 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

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

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

a_{1} + a_{2} + a_{3} + . . . + a_{23} + a_{24} =

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