**Given, **

A man saved in first year, a = 32

A man saved in second year, a_{2 }= 36

Increase saving, d = 4

In n years his saving will be 200, S_{n} = 200

**We know,** S_{n} =\(\frac{n}{2}\) [2a + (n – 1) d]

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

400 = n [64 + 4n – 4]

400 = n [60 + 4n]

400 = 4n [15 + n]

100 = 15n + n^{2}

n^{2} + 15n – 100 = 0

n^{2} + 20n – 5n – 100 = 0

n (n + 20) – 5 (n + 20) = 0

(n – 5) (n + 20) = 0

Here, n – 5 = 0, n = 5

n + 20 = 0,

n = -20

The term can never be negative. So, we consider n = 5

Hence, in 5 years his saving will be Rs. 200