The sum of the first five terms of the series | vedclass

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

Question

(A) The given series is $3 + 4\frac{1}{2} + 6\frac{3}{4} + \dots = 3 + \frac{9}{2} + \frac{27}{4} + \dots$This is a Geometric Progression $(G.P.)$ whe

Vedclass · Accepted Answer

$39\frac{9}{16}$

Vedclass · Answer

(A) The given series is $3 + 4\frac{1}{2} + 6\frac{3}{4} + \dots = 3 + \frac{9}{2} + \frac{27}{4} + \dots$This is a Geometric Progression $(G.P.)$ where the first term $a = 3$ and the common ratio $r = \frac{9/2}{3} = \frac{3}{2}$.The sum of the first $n$ terms of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.For $n = 5$ terms:$S_5 = \frac{3 \left[ (\frac{3}{2})^5 - 1 ight]}{\frac{3}{2} - 1} = \frac{3 \left[ \frac{243}{32} - 1 ight]}{\frac{1}{2}}$$S_5 = 6 \left[ \frac{243 - 32}{32} ight] = 6 \left[ \frac{211}{32} ight] = \frac{3 	imes 211}{16} = \frac{633}{16}$Converting to a mixed fraction: $\frac{633}{16} = 39\frac{9}{16}$.

The sum of the first five terms of the series $3 + 4\frac{1}{2} + 6\frac{3}{4} + \dots$ will be

Similar Questions

Three positive numbers form an increasing $G.P.$ If the middle term in this $G.P.$ is doubled,the new numbers are in $A.P.$ then the common ratio of the $G.P.$ is:

If the sum of the first $6$ terms is $9$ times the sum of the first $3$ terms of the same $G.P.$,then the common ratio of the series will be

Write down the $20^{th}$ term of the $G.P. 1, -1, 1, -1, \ldots$

Five numbers are in $A.P.$,whose sum is $25$ and product is $2520$. If one of these five numbers is $-\frac{1}{2}$,then the greatest number amongst them is

Let $a, b, c \in R$. If $f(x) = ax^2 + bx + c$ is such that $a + b + c = 3$ and $f(x + y) = f(x) + f(y) + xy$ for all $x, y \in R$,then $\sum_{n=1}^{10} f(n)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module