Question

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.

Answer 1

At the end of year 1 man has saved Rs. 32

At the end of year 2 man has saved Rs. 36 + Rs 32 = Rs. 68

At the end of year 1 man has saved Rs. 68 + Rs. 40

The savings form an AP with the first term as 32 common difference as 4 and the $S_n$ as 200

$S_n$ = $\frac{n}{2}$[2a+(n-1)d]

200 = $\frac{n}{2}$[2$\times$32+(n-1)4]

400 = n[60+4n]

Simplifying

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

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

(n+20)(n-5) = 0

n can not be negative

Hence n = 5 years

So it will take him 5 years to save Rs. 200.