A man saved Rs. 32 during the first year, Rs. 36 in the second year and in this way he increases his savings by Rs. 4 every year. Find in what time his saving will be Rs. 200.

A man saved Rs. 32 during the first year, Rs. 36 in the second year and in this way he increases his savings by Rs. 4 every year. To do: We have to find the time in which his savings will be Rs. 200.

Solution:

Amount saved during the first year $=Rs.\ 32$

Savings raised each successive year $=Rs.\ 4$

Let the time in which his savings will be Rs. 200 be $x$ years.

Amount saved each year for $x$ successive years (in Rupees) is,

$32, 36, ......$(x terms)

This is in A.P., where

First term $a = 32$

Common difference $d = 4$

Number of terms $n = x$

We know that,

Sum of $n$ terms in an A.P. $S_n = \frac{n}{2}[2a + (n – 1) d]$

$S_n = \frac{x}{2}[2 (32) + (x – 1) 4]$

$ 200= \frac{x}{2}[64 + 4x-4]$

$2(200) = x(4x+60)$

$400=4x(x+15)$

$100=x^2+15x$

$x^2+15x-100=0$

$x^2+20x-5x-100=0$

$x(x+20)-5(x+20)=0$

$(x+20)(x-5)=0$

$x+20=0$ or $x-5=0$

$x=5$ or $x=-20$ which is not possible

Therefore, in 5 years his savings will be Rs. 200.