Question

A financier lent out a certain sum of money to a person under simple interest at the rate of 4k% p.A, where 'k' is the number of years, i.E. In the first year, the rate is 4%, in the second year, it is 8% and in the third year, it is 12%, and so on. In how many years will his money get doubled?

Answer 1

Given :

Rate of interest = 4k%

where k = number of years

To find :

Number of years in which his money gets doubled.

Solution :

It is given that rate of interest depends on the year

so it means that when is the first year then the rate is 4 %

when is the 2nd year then the rate will become 8% similarly on 3rd year it will become 12 %

then on 4th year 16%

On nth year the rate of interest = 4n%.

and so on ,

so we will calculate the interest on 1st year then interest collected on 2nd year similarly on respective years

so we have to find the number of year at which the total money will become double of the principal amount it means

the interest amount is equal to the principal amount so that the total amount will be double.

So,

Formula of simple interest :

$\\ \bf \: I = \frac{ P \times R \times T}{100}$

where

I = interest

P = Principal amount

R = Rate

T = Time

It is given that rate depends on k

so on first year

rate = 4k = 4%

Time for first year = 1

so interest on first year :

$\\ \bf \: I = \frac{ P \times 4 \times 1}{100}$

on second year

rate = 4k = 8%

Time for second year = 1

so interest on second year :

$\\ \bf \: I = \frac{ P \times 8 \times 1}{100}$

on nth year

rate = 4k = 4n%

Time for nth year = 1

so interest on nth year :

$\\ \bf \: I = \frac{ P \times 4n \times 1}{100}$

It is given that the total amount will become double of the principal amount it means in the interest till nth year is equal to the principal amount at starting .

so, total interest I = P

so

adding interest of all years we get,

$\\ \bf \: P \: = (\frac{ P \times 4 \times 1}{100}) + (\frac{ P \times 8 \times 1}{100}) + ... + (\frac{ P \times 4n \times 1}{100})$

so,

$\\ \bf \: P = (\frac{ 4P }{100}) (1+ 2 + ... + n)$

cancelling P from both sides and multiplying 25 on both sides we get,

$\bf(1 + 2 + 3 + 4 + ... + n) = 25$

we know the formula of sum of n natural numbers as,

$\\ \bf = \frac{(n)(n + 1)}{2}$

so,

The number of years at which the total amount will be doubled will be,

so,

n =(-1-√201)/2=-7.589

n =(-1+√201)/2= 6.589

n can not be negative

so

NUMBER OF YEARS = 6.589