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Question -

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



Answer -

Given:

First year savings is₹ 32

Second year savings is₹ 36

In this process heincreases his savings by ₹ 4 every year

Then,

A.P. will be 32, 36,40,………

Where, 32 is firstterm and common difference, d = 36 – 32 = 4

We know, Sn isthe sum of n terms of an A.P

By using the formula,

Sn =n/2 [2a + (n – 1)d]

where, a is firstterm, d is common difference and n is number of terms in an A.P.

Given:

Sn =200, a = 32, d = 4

Sn =n/2 [2a + (n – 1)d]

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

200 = n/2 [64 + 4n –4]

400 = n [60 + 4n]

400 = 4n [15 + n]

400/4 = n [15 + n]

100 = 15n + n2

n2 +15n – 100 = 0

n2 +20n – 5n – 100 = 0

n (n + 20) – 5 (n +20) = 0

(n + 20) – 5 (n + 20)= 0

(n + 20) (n – 5) = 0

n = -20 or 5

n = 5 [Since, n is apositive integer]

Hence, the manrequires 5 days to save ₹ 200

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