The sum of an infinite number of terms in a G | vedclass

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

Question

(B) Let the first term be $a$ and the common ratio be $r$. The sum of an infinite $G.P.$ is given by $S = \frac{a}{1-r} = 20 \quad (i)$.The sum of the

Vedclass · Accepted Answer

$3/5$

Vedclass · Answer

(B) Let the first term be $a$ and the common ratio be $r$. The sum of an infinite $G.P.$ is given by $S = \frac{a}{1-r} = 20 \quad (i)$.The sum of the squares of the terms is given by $\frac{a^2}{1-r^2} = 100 \quad (ii)$.We can rewrite $(ii)$ as $\frac{a}{1-r} 	imes \frac{a}{1+r} = 100$.Substituting $(i)$ into this equation: $20 	imes \frac{a}{1+r} = 100$,which simplifies to $\frac{a}{1+r} = 5 \quad (iii)$.From $(i)$,$a = 20(1-r)$. Substituting this into $(iii)$:$\frac{20(1-r)}{1+r} = 5$$4(1-r) = 1+r$$4 - 4r = 1 + r$$3 = 5r$$r = 3/5$.

The sum of an infinite number of terms in a $G.P.$ is $20$ and the sum of their squares is $100$. The common ratio of the $G.P.$ is:

Similar Questions

The mean of $10$ numbers $7 \times 8, 10 \times 10, 13 \times 12, 16 \times 14, \ldots$ is ....... .

If the sum of the first $15$ terms of the series ${\left( {\frac{3}{4}} \right)^3} + {\left( {1\frac{1}{2}} \right)^3} + {\left( {2\frac{1}{4}} \right)^3} + {3^3} + {\left( {3\frac{3}{4}} \right)^3} + \dots$ is equal to $225\,k$,then $k$ is equal to

Find the sum to $n$ terms of the series $1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + \ldots$

If $S$ is the sum of $n$ terms of the series $1 + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + \dots$,then $S = \dots$

How many terms are there in the sequence $1, 3, 6, 10, 15, 21, \dots, 5050$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module