Question

A man saves rs 32 during the first year, rs 36 in the second year and in this way he increase his savings by rs 4 every year. find in what time his savings will be rs 200.​

Answer 1

Answer:

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 =[2a + (n – 1) d]

200 =[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

Answer 2

Answer:

use formula

an = a +( n-1)d

a =32

d=4

200=32+(n-1)4

168=(n-1)4

n-1=168/4

n-1=42

n=43

In 43 years....

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