Question

How to solve these type questions​

Answer 1

Answer:

The answer is 7

Step-by-step explanation:

Let  $\text{S}_n=1+3+3^2+3^3+...3^{n-1}$

           $\displaystyle =\frac{1(3^n-1)}{3-1}$, using sum of GP formula

            $\displaystyle =\frac{3^n-1}{2}$

Given  $\text{S}_n>1000$

          $\displaystyle \Rightarrow \frac{3^n-1}{2}>1000\\\\\Rightarrow 3^n>2001$

Also we know $3^6=729,3^7=2187$

So the least value of $n$ is 7

Answer 2

Answer:

• 7

Step-by-step explanation;

Given series is,

${ \small \longrightarrow \underbrace{1 + 3 + {3}^{2} + {3}^{3} + ... + {3}^{n - 1} }_{ \sf Finite\, GP\, with\, a=1\, and \,r=3} > 1000}$

We have formula to find sum of n terms in a GP:

• $\boxed{S_n = \dfrac{a(r^n-1)}{r-1}}$

By using this formula, we get:

${ \small \longrightarrow \dfrac{(1)( {3}^{n} - 1) }{3 - 1} > 1000}$

${ \small \longrightarrow \dfrac{( {3}^{n} - 1) }{2} > 1000}$

${ \small \longrightarrow ( {3}^{n} - 1) > 2000}$

${ \small \longrightarrow {3}^{n} > 2001}$

Now, the smallest multiple of 3 exactly greater than 2001 is 2187 which is 3⁷. Therefore, the required value of n is 7.