Given,
A man saved in first year, a = 32
A man saved in second year, a2 = 36
Increase saving, d = 4
In n years his saving will be 200, Sn = 200
We know, Sn =\(\frac{n}{2}\) [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 + n2
n2 + 15n – 100 = 0
n2 + 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