If in a geometric progression {a_n},a_1 = 3,a | vedclass

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

Question

(B) Given: $a_1 = 3$,$a_n = 96$,and $S_n = 189$.We know that $a_n = a_1 r^{n-1}$,so $3 r^{n-1} = 96$,which implies $r^{n-1} = 32$.Thus,$r^n = 32r$.The

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Given: $a_1 = 3$,$a_n = 96$,and $S_n = 189$.We know that $a_n = a_1 r^{n-1}$,so $3 r^{n-1} = 96$,which implies $r^{n-1} = 32$.Thus,$r^n = 32r$.The sum of $n$ terms of a geometric progression is $S_n = \frac{a_1(r^n - 1)}{r - 1} = 189$.Substituting the values: $\frac{3(32r - 1)}{r - 1} = 189$.Dividing by $3$: $\frac{32r - 1}{r - 1} = 63$.$32r - 1 = 63r - 63$.$62 = 31r$,so $r = 2$.Since $r^{n-1} = 32$,we have $2^{n-1} = 2^5$.Therefore,$n - 1 = 5$,which gives $n = 6$.

If in a geometric progression $\{a_n\}$,$a_1 = 3$,$a_n = 96$ and $S_n = 189$,then the value of $n$ is

Similar Questions

Find the Geometric Mean $(G.M.)$ of the sequence $1, 2, 2^2, \dots, 2^n$.

If the roots of $x^3-42x^2+336x-512=0$ are in increasing geometric progression,then its common ratio is: (in $:1$)

The third term of a geometric progression is equal to the square of the first term. If its second term is $8$,then its sixth term will be:

If $a, b, c$ and $d$ are in $G.P.$,show that:
$(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+d^{2})=(ab+bc+cd)^{2}$

If the $(m + n)^{th}$ term of a geometric progression is $9$ and the $(m - n)^{th}$ term is $4$,what is the $m^{th}$ term?

Vedclass Test Series

Exam Paper Generator

Online Exam Module