The n^{th} term of the series 1 + frac{4}{5} | vedclass

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Question

(C) The given series is an Arithmetico-Geometric progression $(A.G.P.)$.The numerator terms are $1, 4, 7, 10, \dots$,which form an Arithmetic Progress

Vedclass · Accepted Answer

$\frac{3n - 2}{5^{n - 1}}$

Vedclass · Answer

(C) The given series is an Arithmetico-Geometric progression $(A.G.P.)$.The numerator terms are $1, 4, 7, 10, \dots$,which form an Arithmetic Progression $(A.P.)$ with the first term $a = 1$ and common difference $d = 3$.The $n^{th}$ term of this $A.P.$ is $T_n = a + (n - 1)d = 1 + (n - 1)3 = 3n - 2$.The denominator terms are $1, 5, 5^2, 5^3, \dots$,which form a Geometric Progression $(G.P.)$ with the first term $a = 1$ and common ratio $r = 5$.The $n^{th}$ term of this $G.P.$ is $G_n = ar^{n - 1} = 1 \cdot 5^{n - 1} = 5^{n - 1}$.Therefore,the $n^{th}$ term of the given series is the ratio of the $n^{th}$ term of the $A.P.$ to the $n^{th}$ term of the $G.P.$,which is $\frac{3n - 2}{5^{n - 1}}$.

The $n^{th}$ term of the series $1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + \dots$ will be

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