# Jaspal Singh repays his total loan of Rs 118000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, what amount will be paid by - Mathematics

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Jaspal Singh repays his total loan of Rs 118000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, what amount will be paid by him in the 30th instalment? What amount of loan does he still have to pay after the 30th instalment?

#### Solution

Given that,

Jaspal singh takes total loan = ₹ 118000

He repays his total loan by paying every month

His first instalment = ₹ 1000

Second instalment = 1000 + 100 = ₹ 1100

Third instalment = 1100 + 100 = ₹ 1200 and so on

Let its 30th instalment be n,

Thus, we have 1000,1100,1200,… which form an AP, with first term (a) = 1000

and common difference (d) = 1100 – 1000 = 100

nth term of an AP, Tn = a + (n – 1)d

For 30th instalment, T30 = 1000 + (30 – 1)100

= 100 + 29 × 100

= 1000 + 2900

= 3900

So, ₹ 3900 will be paid by him in the 30th instalment.

He paid total amount upto 30 instalments in the following form

1000 + 1100 + 1200 + …………. + 3900

First-term (a) = 1000 and last term (1) = 3900

∴ Sum of 30 instalments, S_30 = 30/2 [a + 1]

∵ sum of first n terms of an AP is S_n = n/2 [a + 1]

Where l = last term

⇒ S30 = 15(1000 + 3900) = 15 × 4900 = ₹ 73500

⇒ Total amount he still have to pay after the 30 installment

= (Amount of loan) – (Sum of 30 installments)

= 118000 – 73500

= ₹ 44500

Hence, ₹ 44500 still have to pay after the 30th installment.

Concept: Sum of First n Terms of an A.P.
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#### APPEARS IN

NCERT Mathematics Exemplar Class 10
Chapter 5 Arithematic Progressions
Exercise 5.4 | Q 9 | Page 56
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