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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

$\left[\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{99 \times 100}\right]$ is equal to

$\frac{2}{5} + \frac{3}{5^{2}} + \frac{2}{5^{3}} + \frac{3}{5^{4}} + \dots \infty$

The $3^{rd}$ term of a $G.P.$ is the square of the first term. If the second term is $8,$ determine the $6^{th}$ term.

If five $G.M.s$ are inserted between $486$ and $2/3$,then the fourth $G.M.$ will be:

If the sum of the first $40$ terms of the series $3+4+8+9+13+14+18+19+\ldots$ is $(102)m$,then $m$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module