The sum of some terms of a G.P. is 315,whose | vedclass

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

Question

(A) Let the sum of $n$ terms of the $G.P.$ be $S_n = 315$.The formula for the sum of $n$ terms of a $G.P.$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.Given t

Vedclass · Accepted Answer

Last term: $160$,Number of terms: $6$

Vedclass · Answer

(A) Let the sum of $n$ terms of the $G.P.$ be $S_n = 315$.The formula for the sum of $n$ terms of a $G.P.$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.Given that the first term $a = 5$ and the common ratio $r = 2$.Substituting the values into the formula: $315 = \frac{5(2^n - 1)}{2 - 1}$.$315 = 5(2^n - 1)$.$2^n - 1 = \frac{315}{5} = 63$.$2^n = 64 = 2^6$.Therefore,$n = 6$.The last term of the $G.P.$ is the $n^{th}$ term,given by $a_n = a \cdot r^{n-1}$.$a_6 = 5 \cdot 2^{6-1} = 5 \cdot 2^5 = 5 \cdot 32 = 160$.Thus,the last term is $160$ and the number of terms is $6$.

The sum of some terms of a $G.P.$ is $315$,whose first term and the common ratio are $5$ and $2$ respectively. Find the last term and the number of terms.

Similar Questions

For a sequence,$a_1 = 2$ and $\frac{a_{n+1}}{a_n} = \frac{1}{3}$. Then $\sum_{r=1}^{20} a_r$ is

If the first term of a $G.P.$ is $5$ and the common ratio is $-5$,then which term is $3125$ (in $^{th}$)?

If $x = \sum\limits_{n = 0}^\infty {{a^n}} ,\;y = \sum\limits_{n = 0}^\infty {{b^n},\;z = \sum\limits_{n = 0}^\infty {{{(ab)}^n}} } $,where $a, b < 1$,then

If $1 + r + r^2 + \dots + r^n = (1 + r) (1 + r^2) (1 + r^4) (1 + r^8)$,then what is the value of $n$?

What is the geometric mean of the observations $2, 4, 8, 16, 32, 64$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module