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

  • Option 1)

     

    4(5^{2})

  • Option 2)

     

    5^{4}

  • Option 3)

     

    2(5^{2})

  • Option 4)

     

    5^{3}

Answers (1)

best_answer

 

Geometric Progession (GP) -

A progression of non - zero terms, in which every term bears to the preceding term a constant ratio.

- wherein

eg 2, 4, 8, 16,- - - - - -

and

100, 10, 1, 1/10,- - - - - - -

 

 

General term of a GP -

T_{n}= ar^{n-1}
 

- wherein

a\rightarrow first term

r\rightarrow common ratio

 

Let first term of G.D be a and common ratio = r

a_n=ar^{n-1}

\frac{a_3}{a_1}=\frac{ar^{2}}{a}=25\Rightarrow r^{2}=25

Now, \frac{a_9}{a_5}=\frac{ar^{8}}{ar^{4}}=r^{4}=\left ( 25 \right )^{2}=5^{4}


Option 1)

 

4(5^{2})

Option 2)

 

5^{4}

Option 3)

 

2(5^{2})

Option 4)

 

5^{3}

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