Question

A man saves 135/- in the first year, 150/- in the second year and in this way he increases his savings by 15/ every year. In what time will his total savings be 5550/-?
20years
25years
30years
35years

Answer 1

since there is increase of 15 every year,

this is an arithmetic sequence

135,150,165....

first term is 135

common difference is 15

sum is 5550

s=n/2(2a+(n-1)d)

5550=n/2[(2×135)+(n-1)×15]

5550=n/2[270+15n-15]

11100=n(255+15n)

1110=255n+15n²

15n²+255n+11100=0

divide by 15

n²+17n+740=0

n²+37n-20n+740=0

n(n+37)-20(n+37)=0

(n-20)(n+37)=0

n=20 or -37

Since year cannot be negative

n=20 years

Answer 2

Given: A man saves 135/- in the first year, 150/- in the second year and in this way he increases his savings by 15/- every year.

To find: Time in which his total savings will be 5550/-

Explanation: Since the man increases his savings by 15/- every years, the sequence of savings form an arithmetic progression. In this AP, the first term is equal to the amount invested in first year and the common difference is the money increased every year.

Let the first term be a and common difference be d.

a = 135

d = 15

Let after n years his total savings his 5550. Therefore, the sum of money invested should be 5550 after n years.

Sum of AP(s) is given by:

$s = \frac{n}{2} (2a + (n - 1)d)$

$= > 5550 \times 2 = 2na + n(n - 1)d$

$= > 11100 = 2 \times 135 \times n + ({n}^{2} - n)15$

$= > 11100 = 270n + 15 {n}^{2} - 15n$

$= > 15 {n}^{2} +255n -11100 = 0$

$= > {n}^{2} +17n - 740 = 0$ (dividing by 15)

$= > {n}^{2} + 37n -20n - 740 = 0$

$= > n(n + 37) - 20(n + 37) = 0$

$= > (n + 37)(n - 20) = 0$

=> n+37 = 0 or n-20 = 0

=> n = -37 or n = 20

But number of years cannot be negative, therefore n= 20

Therefore, after 20 years the man's total savings will be 5550/-.