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.
Given:
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.
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