Question

Jaspal Singh repairs is total loan of Rupees 1 18000 by paying every month starting with the first installment of Rs.1000 if we increase the installment by Rs.100 every month what amount will be paid by him in the 30th installment ? what amount of loan does he still have to pay after the 30th installment?​

Answer 1

Answer:

Let us assume that the loan is cleared in 'n' months.

Then, The amounts are in AP with first term (a) = 1000, the common difference (d) = 100 and the Sum of amounts (Sn) = Rs. 118000

So, according to the question.

Sn = n/2{2a + (n - 1)d}

118000 = n/2{2 × 1000 + (n - 1)100}

236000 = n{2000 - 100 + 100n}

236000 = 1900n + 100n²

100n² + 1900n - 236000 = 0

Dividing the above by 100, we get.

n² + 19 - 2360 = 0

n² + 59n - 40n - 2360 = 0

n(n + 59) - 40(n +59) = 0

(n - 40) (n + 59) = 0

n = 40 or n = -59

Since the value of cannot be negative so, n = 40 is correct.

Therefore, the loan will be cleared in 40 months.

Now, the amount to be paid by him in the 30th installment is calculated as under.

a₃₀ = a + 29d  where a = 1000 and d = 100

a₃₀ = 1000 + (29*100)

= 1000 + 2900

a₃₀ = 3900

so, the amount to be paid by him in the 30th installment is Rs. 3900

Now, the amount of remaining loan that he has to pay after the 30th installment is calculated as under.

S₃₀ = n/2{2a + (n - 1)d}

= 30/2{2 × 1000 + (30 - 1)100}

= 15{2000 + 2900}

= 15*4900

= Rs. 73500

So, the total amount paid after the 30th installment is Rs. 73500.

The remaining amount of loan = total loan amount - amount paid after the 30th installment

= 118000 - 73500

= Rs. 44500 

So, the he still has to pay Rs. 44500 after the payment of the 30th installment.