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Question
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?
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Solution
Given that,
Jaspal singh takes total loan = Rs. 118000
He repays his total loan by paying every month
His first instalment = Rs. 1000
Second instalment = 1000 + 100 = Rs. 1100
Third instalment = 1100 + 100 = Rs. 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
= 1000 + 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 (l) = 3900
∴ Sum of 30 instalments,
S30 = `30/2 [a + l]` ...[∵ Sum of first n terms of an AP is, `S_n = n/2 [a + l]` where l = last term]
⇒ S30 = 15(1000 + 3900)
= 15 × 4900
= Rs. 73500
⇒ Total amount he still have to pay after the 30th installment
= (Amount of loan) – (Sum of 30 installments)
= 118000 – 73500
= Rs. 44500
Hence, Rs. 44500 still have to pay after the 30th installment.
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