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

#### Solution

Here, we are given that the total saving of a man is Rs 200. In the first year, he saved Rs 32 and every year he saved Rs 4 more than the previous year.

So, the first instalment = 32*.*

`Second installment = 36

Third installment = 36 + 4

So, these installments will form an A.P. with the common difference (*d*) = 4

The sum of his savings every year `S_n = 200`

We need to find the number of years. Let us take the number of years as *n.*

So, to find the number of years, we use the following formula for the sum of *n *terms of an A.P.,

`S_n = n/2[2a + (n - 1)d]`

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, using the formula for *n* = 10, we get,

`S_n = n/2[2(32) + (n -1)(4)]`

`200 = n/2 [64 + 4n - 4]`

400 = n(60 + 4n)

`400 = 60n + 4n^2`

We get a quadratic equation,

`4n^2 + 60n - 400 = 0`

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

Further solving for *n* by splitting the middle term, we get,

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

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

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

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

So,

n - 5 = 0

n = 5

Or

n + 20 = 0

n = -20

Since number of years cannot be negative. So in 5 years, his savings will be Rs 200