Question

Jaspal Singh repays his total loan of Rs.118000 by paying every month starting with the first installment of Rs.1000. If he increases the installment by Rs.100 every month, what amount will be paid by him in ${{30}^{th}}$ installment?

Answer 1

Hint: We can solve this question by writing the given condition in the form of A.P because the amount is increasing every month by a fixed amount. That’s why A.P will be formed according to given conditions. Then we can find ${{30}^{th}}$ term of that A.P.

Complete step-by-step solution -
Total loan of Jaspal Singh is Rs.118000.
First installment of loan paid by Jaspal is Rs.1000
Every month installment is increasing Rs. 100
Hence we can write it as $1000,1100,1200,1300,.......................$
So in above A.P first term(a) = 1000
Common difference(d) = 100
For finding term of A.P we can use general term formula of A.P i.e ${{t}_{n}}=a+(n-1)d$ ………………………………(i)
Where a is first term, d is common difference and n is number of terms.
On substituting value of a , d and n
$\Rightarrow {{t}_{30}}=1000+(30-1)\times 100$
$\Rightarrow {{t}_{30}}=1000+29\times 100$
$\Rightarrow {{t}_{30}}=1000+2900$
$\Rightarrow {{t}_{30}}=3900$
Hence the amount paid in ${{30}^{th}}$ installment of loan is Rs. 3900.

Note: A series is an A.P if there is a common difference between two consecutive terms of series. We can write an A.P if first term is a and common difference is d as below:
$a,a+d,a+2d,a+3d,.................$
In above pattern we can write general term of an A.P as
${{t}_{n}}=a+(n-1)d$.