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If the general term, Tn = \(\rm \frac{5^n}{2^{n -1}}\), then what is the value of \(\rm 1st \ term\ \over Common \ ratio\)

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Correct Answer - Option 3 : 2

Concept:

For a sequence a1, a2, a3 …. an is an G.P.

Common ratio , r = \( \rm a_2 \over a_1\) = \( \rm a_3 \over a_2\) = \( \rm a_n \over a_{n-1}\)

Calculation:

The given general term is,

Tn = \(\rm \frac{5^n}{2^{n -1}}\)

Putting the value of n= 1, 2, 3... we get,

T1 = 5

T2 = \(\rm 5^2 \over 2^{2-1}\) = \(25\over2\)

T3 = \(\rm 5^3 \over 2^{3-1}\) = \(125\over4\)

So the G.P series becomes, 5, \(25\over2\)\(125\over4\)

Where the 1st term is 5 and common ratio is \(25\over2\ \times\ 5\) = \(5\over2\)

So, \(\rm 1st \ term\ \over Common \ ratio\) = \(\rm \frac{5}{\frac{5}{2}}\)= 2

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