Question

A man repays a loan of Rs. 3250 by paying Rs. 20 in the first month and then increases the payment by Rs 15 every month.How long will it take him to clear the loan?

Answer 1

`20` `"months"`

Explanation:

In order to solve or understand a question like this, we need to know about arithmetic sequences.

Arithmetic sequences are the sequences which have a constant difference between the numbers. In other words, it is the sequence of numbers in which we obtain the next term by adding a constant number.So, it is in the form of `(color(purple)(a,a+d,a+2d,a+3d....)`

For example,

`color(purple)(|->3,5,7,9,11,13...`

Is an arithmetic sequence. In this the common difference is `color(orange)(2`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now, let's understand this question. In this, he pays `3250` by giving `20` in the first month. And then, increases it by `15` every month thereafter.

`color(violet)(|->20,35,50,65....`

In this manner he clears the loan in `n` months

To solve this, we need to know the formula

`color(brown)(S_n=n/2(2a+(n-1)d)`

Where, `S_n` is the sum of `n` number of terms. `a` is the first term(`20`). And, `d` is the common difference (In this case, it is `15`). Finally, `n` is the number of terms

According to the question, We need to find `n`. And also, they have given `S_n=3250`

So,

#rarrS_n=n/2(2a+ (n-1)d)#

`rarr3250=n/2(2(20)+(n-1)(15))`

`rarrn(40+15n-15)=2xx3250`

`rarrn(25+15n)=6500`

`rarr15n^2+25n-6500=0`

`rarr3n^2+5n-1300=0`

`rarr(n-20)(3n+65)=0`

`rarrn-20=0`

`color(green)(rArrn=20` `color(green)("months"`

Hope that helps!...`phi`