A man purchased a home with the

Question

help of a bank loan. He plans to pay
back the entire debt of Rs. 3,30,600
in monthly instalments, beginning
with the first instalment of Rs. 5,000.
He then decided to increase the
instalment amount every month by
200 which forms an AP. It will take
him 38 months to repay the loan in
this manner.
Based on the above
information, answer the following
questions
(i) How much money will he pay in her 25 instalment?
(ii) In which month he will pay rs. 8000 to
the bank as an instalment?
(iii) How much money would be paid till the 30th instalment to the bank?
(iv)In how many months he will pay rs. 1,38,000 to the bank?

Answer 1

not familiar with your notation of 3,30,600
at first I though it was a typo and you meant 330,000, and used it
in my calculation. Then I saw it again at the end as 1,38,000.
Please explain.

i)
The payments consist of the AP
5000, 5200, 5400, ...
a = 5000, d = 200
25th installment = term(25) = a + 24d = .....

ii) term(n) = 8000
a + (n-1)d = 8000
5000 + 200n - 200 = 8000
200n = 3200
n = 16
the 16th month

iii) This is an invalid question, since you can't just add amounts of money
that are at different time spots.
simpleton answer: sum of 5000 + 5200 + 540 + ... for 30 terms
= (30/2)(10000 + 29(200)) = 237,000

iv) again, not a valid question, unless there is no interest involved.
simpleton answer, assuming 0% interest:
138,000 = (n/2)(10,000 + 200(n-1)
276,000 = 10,000n + 200n^2 - 200n
200n^2 + 9800n - 276000 = 0
n^2 + 49n - 1380 = 0

solve as a quadratic in n , hint: it factors, reject the negative answer

(if you had given an interest rate, this would become a very interesting
problem)

Answer 2

You have done AP exercises before -- what's the problem with this one?
The nth payment,
p_n = 5000+200(n-1) = 4800+200n = 200(24+n)
(i) p_25 = 200(24+25)
(ii) find n where 200(24+n) = 8000
(iii) 30/2 (p_1 + p_30)
(iv) find n such that n/2 (2*5000 + 200(n-1)) = 138,000