# Case study of maths

Dear student,
Here, Total amount of loan =1,18,000
Since the amount of each installment increases by Rs.100 every month.
∴ installments paid are in AP.
a) Amount of first installment a=Rs.1000
Increase in amount of each installment d=Rs.100
Option (i) is correct.

b) S​​​​​​n = 118000 = (n/2)[2×a+(n-1)d]
118000 = (n/2)[2×1000+(n-1)100]
On solving, we get n = 40
Option (iii) is correct.

c) Amount paid by jaspal singh in 30th installment=a​​​​​30
Now, a​​​​​​30 =a + 29d =1000+29×100 = Rs.3900
Option (i) is correct.

d) Amount paid in 30 installments,
= (30/2)[2×1000+(30−1)×100]
=30 [2450]
= Rs. 73500
Option (iii) is correct.

e) Amount of loan still paid by jaspal after 30 installments
= Total loan-amount paid in 30 installments
= 118000 - $\frac{n}{2}$[2×1000+(n-1)100]
=118000− $\frac{30}{2}$ [2×1000+(30−1)×100]
=118000−30[2450]
=118000−73500
=Rs.44500
Option (iv) is correct.

Regards.

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