Madhya Pradesh Police Exam Model Paper 4

Concept:
Let us consider sequence ${\mathit{a}}_1,{\mathit{a}}_2,{\mathit{a}}_3\dots .{\mathit{a}}_{\mathit{n}}$ is a G.P.

• Common ration = r= $\frac{{\mathit{a}}_2}{{\mathit{a}}_1}=\frac{{\mathit{a}}_3}{{\mathit{a}}_2}=.....\frac{{\mathit{a}}_{\mathit{n}}}{{\mathit{a}}_{\mathit{n}-1}}$
  
• ${\mathit{n}}^{\mathit{t}\mathit{h}}$ term of the G.P. is ${\mathit{a}}_{\mathit{n}}=\mathit{a}{\mathit{r}}^{\mathit{n}-1}$
• Sum of n terms of GP = ${\mathit{S}}_{\mathit{n}}=\mathit{a}\frac{\left({\mathit{r}}^{\mathit{n}}-1\right)}{\mathit{r}-1}$; where $\mathit{r}>1$
• Sum of n terms of GP= ${\mathit{S}}_{\mathit{n}}=\mathit{a}\frac{\left(1-{\mathit{r}}^{\mathit{n}}\right)}{\mathit{r}-1}$; where
• Sum of infinite GP $={\mathit{S}}_{\mathit{\infty }}=\frac{\mathit{a}}{1-\mathit{r}}$;

Calculation:
Given series is 5, 10, 20, ...
Here, a = 5, r = 2
Sum of n numbers $={\mathit{S}}_{\mathit{n}}=1275$
To Find: nAs we know that, Sum of n terms of GP $={\mathit{S}}_{\mathit{n}}=\frac{{\mathit{r}}^{\mathit{n}}-1}{\mathit{r}-1}$ ;where $\mathit{r}>1$
$\therefore {\mathit{S}}_{\mathit{n}}=\frac{5\left(2^{\mathit{n}}-1\right)}{2-1}$
$1275=5\times 2^{\mathit{n}}-1$
$\Rightarrow 255=2^{\mathit{n}}-1$
$\Rightarrow 2^{\mathit{n}}=256$
$\Rightarrow 2^{\mathit{n}}=2^8$
$\therefore \mathit{n}=8$