Find the sum of the G.P.: 2/5, 2/25, 2/125, 2/625, … to the n terms.

Find the sum of the G.P.:
2/5, 2/25, 2/125, 2/625, … to the n terms.

a) 5/4(1-(1/5ⁿ))
b) 2/5(1-(1/5ⁿ))
c) 1/2(1-(1/5ⁿ))
d) 4/5(1-(1/5ⁿ))

Answer : c) 1/2(1-(1/5ⁿ))

UP Police SI Paper – Numerical Mental Ability

Solution :

To find the sum of the given geometric progression (G.P.), we need to identify the first term and the common ratio.
The first term ( a ) is 2/5
The common ratio ( r ) is found by dividing the second term by the first term: $r = \frac{\frac{2}{25}}{\frac{2}{5}} = \frac{2}{25} \times \frac{5}{2} = \frac{1}{5}$

The sum of the first ( n ) terms of a G.P. is given by the formula: $S_n = a \frac{1 - r^n}{1 - r}$

Substituting the values of ( a ) and ( r ): $S_n = \frac{2}{5} \frac{1 - \left(\frac{1}{5}\right)^n}{1 - \frac{1}{5}}$

$S_n = \frac{2}{5} \times \frac{5}{4} \times \left(1 - \left(\frac{1}{5}\right)^n\right)$
$S_n = \frac{2}{4} \left(1 - \left(\frac{1}{5}\right)^n\right)$
$S_n = \frac{1}{2} \left(1 - \left(\frac{1}{5}\right)^n\right)$

So, the sum of the first ( n ) terms of the given G.P. is: $S_n = \frac{1}{2} (1 - (\frac{1}{5})^n) =\frac{1}{2} (1 - (\frac{1^n}{5^n})) =\frac{1}{2} (1 - (\frac{1}{5^n}))$