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

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

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

Solution:

As per the question:
Savings for the first year is ₹ 32
Savings for the second year is ₹ 36
Every year he increases his savings by ₹ 4.
Therefore,
A.P. will be $32,36,40, \ldots \ldots \ldots$
Where, the first term is 32 and the common difference, $d=36-32=4$
It is known that, $S_{n}$ is the sum of $n$ terms of an A.P
Using the formula,
$S_{n}=n / 2[2 a+(n-1) d]$
Where, the first term is $a$ , the common difference is $d$ and $n$ is number of terms in an A.P.
Given that,
$S_{n}=200, a=32, d=4$
$\begin{array}{l} S_{n}=n / 2[2 a+(n-1) d] \\ 200=n / 2[2(32)+(n-1) 4] \\ 200=n / 2[64+4 n-4] \\ 400=n[60+4 n] \\ 400=4 n[15+n] \\ 400 / 4=n[15+n] \\ 100=15 n+n^{2} \\ n^{2}+15 n-100=0 \\ n^{2}+20 n-5 n-100=0 \\ n(n+20)-5(n+20)=0 \\ (n+20)-5(n+20)=0 \\ (n+20)(n-5)=0 \\ n=-20 \text { or } 5 \end{array}$
$n=5 \text { [Since, } n \text { is a positive integer] }$
As a result, the man requires 5 days to save ₹200