# Case study of maths

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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