Let the sum of an infinite G.P.,whose first t | vedclass

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

Question

(A) Given the sum of an infinite $G.P.$ is $S_{\infty} = \frac{a}{1-r} = 5$,so $a = 5(1-r)$.The sum of the first five terms is $S_5 = \frac{a(1-r^5)}{

Vedclass · Accepted Answer

$21 a_{11}$

Vedclass · Answer

(A) Given the sum of an infinite $G.P.$ is $S_{\infty} = \frac{a}{1-r} = 5$,so $a = 5(1-r)$.The sum of the first five terms is $S_5 = \frac{a(1-r^5)}{1-r} = 5(1-r^5) = \frac{98}{25}$.$1-r^5 = \frac{98}{125} \implies r^5 = 1 - \frac{98}{125} = \frac{27}{125} = (\frac{3}{5})^5$,so $r = \frac{3}{5}$.Then $a = 5(1 - \frac{3}{5}) = 5(\frac{2}{5}) = 2$.For the $A.P.$,the first term $A = 10ar = 10(2)(\frac{3}{5}) = 12$ and the common difference $D = 10ar^2 = 10(2)(\frac{9}{25}) = \frac{36}{5} = 7.2$.The sum of the first $21$ terms is $S_{21} = \frac{21}{2} [2A + (21-1)D] = \frac{21}{2} [2(12) + 20(7.2)] = \frac{21}{2} [24 + 144] = \frac{21}{2} [168] = 21 	imes 84 = 1764$.Now,calculate $a_{11}$ of the $A.P.$: $a_{11} = A + 10D = 12 + 10(7.2) = 12 + 72 = 84$.Thus,$S_{21} = 21 	imes 84 = 21 a_{11}$.

Similar Questions

Let $I(n) = n^n$ and $J(n) = 1 \times 3 \times 5 \times \ldots \times (2n - 1)$ for all $n > 1, n \in N$. Then:

If $t_n$ is the $n^{th}$ term of an arithmetic progression and $t_7 = 9$,what is the value of the common difference $d$ that minimizes the product $t_1 t_2 t_7$?

If $a, b, c$ are in $G.P.$ and $x, y$ are the arithmetic means between $a, b$ and $b, c$ respectively,then $\frac{a}{x} + \frac{c}{y}$ is equal to

Two sequences $\{t_n\}$ and $\{s_n\}$ are defined by $t_n = \log \left( \frac{5^{n+1}}{3^{n-1}} \right)$ and $s_n = \left[ \log \left( \frac{5}{3} \right) \right]^n$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module