How many terms of G.P. 3,3^{2}, 3^{3}... are | vedclass

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

Question

The given $G.P.$ is $3,3^{2}, 3^{3} \ldots$

Let $n$ terms of this $G.P.$ be required to obtain in the sum as $120 .$

$S_{n}=\frac{a\left(1-r^{n}

Vedclass · Accepted Answer

Vedclass · Answer

The given $G.P.$ is $3,3^{2}, 3^{3} \ldots$

Let $n$ terms of this $G.P.$ be required to obtain in the sum as $120 .$

$S_{n}=\frac{a\left(1-r^{n}ight)}{1-r}$

Here, $a=3$ and $r=3$

$	herefore S_{n}=120=\frac{3\left(3^{n}-1ight)}{3-1}$

$\Rightarrow 120=\frac{3\left(3^{n}-1ight)}{2}$

$\Rightarrow \frac{120 	imes 2}{3}=3^{n}-1$

$\Rightarrow 3^{n}-1=80$

$\Rightarrow 3^{n}=81$

$\Rightarrow 3^{n}=3^{4}$

$	herefore n=4$

Thus, four terms of the given $G.P.$ are required to obtain the sum as $120 .$

How many terms of $G.P.$ $3,3^{2}, 3^{3}$... are needed to give the sum $120 ?$

Similar Questions

The sum of first three terms of a $G.P.$ is $\frac{39}{10}$ and their product is $1 .$ Find the common ratio and the terms.

In an increasing geometric progression ol positive terms, the sum of the second and sixth terms is $\frac{70}{3}$ and the product of the third and fifth terms is $49$. Then the sum of the $4^{\text {th }}, 6^{\text {th }}$ and $8^{\text {th }}$ terms is :-

Which term of the $GP.,$ $2,8,32, \ldots$ up to $n$ terms is $131072 ?$

Suppose that the sides $a,b, c$ of a triangle $A B C$ satisfy $b^2=a c$. Then the set of all possible values of $\frac{\sin A \cot C+\cos A}{\sin B \cot C+\cos B}$ is

$2.\mathop {357}\limits^{ \bullet \,\, \bullet \,\, \bullet } = $