Question

A man saves Rs 135 in first year, Rs150in the second year and in this way he increases his savings by Rs 15 every year. Inwhat time will his totat be Rs5550

Answer 1

Given:

In first year a man saved, a = Rs 135

In second year a man saved Rs 150

Increase in savings, d = Rs 15

His saving in 'n' years will be , Sn = 5550  

To find :

In How many years will his total be Rs 5550

Formula:

$S_n = \frac{n}{2}[2a+(n-1)d]$

Solution:

Step 1 of 2:

Let 'n' be the number of years,

$S_n = \frac{n}{2}[2a+(n-1)d]$

$S_n = \frac{n}{2}[2(135)+(n-1)15]$

$S_n = 5550$

5550 = $\frac{n}{2}[270+(n-1)15]$

5550 × 2 = n[270 + 15n - 15]

11100 = n[255 + 15n]

11100 = 255n + 15n²

15n² + 255n - 11100 = 0

On dividing by 15 we get,

n² + 17n - 740 = 0

Step 2 of 2:

n² + 17n - 740 = 0

By splitting the middle terms,

n² +  37n - 20n - 740 = 0

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

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

n- 20 = 0

n = 20

n + 37 = 0

n = -37 Number of years cannot be negative.

Therefore, The number of years is 20.

His saving in 20 years will be Rs 5550

Final answer:

In 20 years his total will be Rs 5550

Answer 2

Series and sequence

Given:

Year by year savings are depicted, total savings of 5550.

To Find:  

Years taken to accumulate savings of 5550.

Explanation:

In this question, first year savings was 135, 2nd years savings increased with 15 and became 150 and so on.

Let us suppose the number of years taken to achieve a saving of 5550, means sum of savings should be 5550 = ' n' years

So, we can see pattern of it as,

$135, 150, 165,\_,\_,\_,_,\_,\_,_,\_,\_,\_\_\_n$

So, sum of this series =$5550 = S_{n}$

Difference between each savings$= 15 = d$

1st term $= 135 = a$

The formula that we will use is,

$S_{n}=\frac{n}{2}(2a+(n-1)d)$, as this series makes an arithmetic progression series.

Put the values in this formula,

$5550=\frac{n}{2}(2\times135+(n-1)15)\\\\5550=\frac{n}{2}(270+15n-15)\\\\5550=\frac{n}{2}(255+15n)\\\\11100=15n^{2}+255n\\\\ 15n^{2}+255n-11100=0\\\\n^{2}+17n-740=0\\\\(n+37)(n-20)=0\\\\n=-37 \\n= 20$

we reject -37 .

Hence ,in  20 years the savings will be 5550.