Question

a person deposits rs. 1,000 in the first month. then every month he increases the monthly deposit by rs. 60. use the principle of progression and calculate his total investment at the end of two years. find its solution

Answer 1

given that,
a=first term=1000
r=increase in the monthly deposit=60
we know for an A.P sequence, sum upto n terms=n[a+(n-1)d]/2
so as we have 1 year= 12 months => 2 yrs = 24 months,
  we get n=24
hence total investment at the end of 2 years is given by,
   =>     n[a+(n-1)d]/2
   =>     24[1000+23*60]/2
   =>     24[2380]/2
   =>     12*2380
   =>     28560

Answer 2

His total investment after two years is Rs. 2380.

Given,

A person deposits Rs. 1,000 in the first month. Then every month he increases the monthly deposit by rs. 60.

To Find,

His total investment at the end of two years.

Solution,

The method of finding his total investment at the end of two years is as follows -

We know when an amount is increased by a fixed value after a certain amount of time again and again, it creates an Arithmetic progression.

If the first term of an AP is 'a' and the common difference is 'd', the nth term of that AP will be $a+(n-1)d$.

Here in the given AP, the first term a = Rs. 1000 and the common difference is d = Rs. 60. The total time lapse is 2 years = 24 months.

So the total amount of investment is $a+(n-1)d=1000+(24-1)*60=1000+23*60$

$=1000+1380=2380$.

Hence, his total investment after two years is Rs. 2380.

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