A man purchased a home with the
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?
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)
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
To find the answers to these questions, we need to analyze the given information step by step.
First, let's understand the pattern of the monthly installment amounts. The man starts with an installment of Rs. 5,000 and increases it by Rs. 200 every month. This means that the installment amounts form an arithmetic progression (AP).
(i) To calculate the amount the man will pay in the 25th installment, we need to find the value of the 25th term of the AP. The formula for finding the nth term of an AP is given by:
an = a + (n-1)d
Where:
an = nth term
a = first term
n = number of terms
d = common difference
In this case, the first term (a) is Rs. 5,000 and the common difference (d) is Rs. 200.
Plugging in these values, we can calculate the 25th term:
a = 5000
d = 200
n = 25
an = 5000 + (25-1)200
= 5000 + 24*200
= 5000 + 4800
= 9800
Therefore, the man will pay Rs. 9,800 in the 25th installment.
(ii) To find the month in which the man will pay Rs. 8000, we need to find the term number of this amount. We can use the same process as in part (i) but set the value of the installment amount (an) to Rs. 8000 and solve for n.
a = 5000
d = 200
an = 8000
8000 = 5000 + (n-1)200
3000 = (n-1)200
15 = n-1
n = 16
Therefore, the man will pay Rs. 8000 to the bank in the 16th month.
(iii) To determine how much money the man would have paid to the bank till the 30th installment, we need to find the sum of the terms from the first term to the 30th term. The sum of the first n terms of an AP is given by:
Sn = (n/2)(2a + (n-1)d)
Plugging in the values, we can calculate the sum till the 30th installment:
a = 5000
d = 200
n = 30
Sn = (30/2)(2*5000 + (30-1)*200)
= 15(10000 + 29*200)
= 15(10000 + 5800)
= 15(15800)
= 237000
Therefore, the man would have paid Rs. 2,37,000 to the bank till the 30th installment.
(iv) Lastly, to determine in how many months the man will pay Rs. 1,38,000, we need to find the term number when the sum of all installments is equal to or exceeds Rs. 1,38,000.
Using the same formula as in part (iii), we can equate Sn to 1,38,000 and solve for n:
a = 5000
d = 200
Sn = 138000
138000 = (n/2)(2*5000 + (n-1)*200)
27600 = n(10000 + 200n - 200)
27600 = 10000n + 200n^2 - 200n
200n^2 + 9800n - 27600 = 0
We can solve this quadratic equation using factoring, completing the square, or using the quadratic formula. Solving it, we find that n ≈ 13.8 (approximately).
Since the number of months cannot be a fraction, we round it up to the next whole number. Hence, the man will pay Rs. 1,38,000 to the bank in the 14th month.
To summarize:
(i) The man will pay Rs. 9,800 in the 25th installment.
(ii) The man will pay Rs. 8,000 to the bank in the 16th month.
(iii) The man would have paid Rs. 2,37,000 to the bank till the 30th installment.
(iv) The man will pay Rs. 1,38,000 to the bank in the 14th month.