Calculation of Interest on Recurring Deposit –
Suppose $ P per month is deposited each month for ‘n’ months at R% per annum. Then, $ P deposited in the ‘n’th month will earn interest for 1 month, that deposited in (n – 1)th month will earn interest for 2 months, and so on, while the sum deposited in the first month will earn interest for ‘n’ months.
Thus we have –
Equivalent principal for one month = $ [P X (1 + 2 + 3 + ……………. + n)]
n(n + 1)
= $ [P X -----------]
2
Thus, the interest can be calculated using the formula
n(n + 1) 1 R
S.I. = $ [P X ----------- X ------- X -------]
2 12 100
Maturity value (M.V) = (P X n) + I
There are some examples are given below for your better understanding -
Example.1) Richard deposited $ 200 per month for 36 months in a bank’s recurring deposit account. If the bank pays interest at 12% per annum, find the amount she gets on maturity.
Ans.) Here P = $ 200, n = 36 months, R = 12% p.a
Amount deposited in 36 months = $ (200 X 36) = $ 7200
n(n + 1) 1 R
S.I. = $ [P X ----------- X ------- X -------]
2 12 100
1 1 12
= $ [200 X ------- X 36 X (36 + 1) X ------- X -------]
2 12 100
1
= $ [100 X 36 X 37 X -------]
100
= $ (36 X 37) = $ 1332
The amount received by Richard on maturity = $ (1332 + 7200)
= $ 8532 (Ans.)
Example.2) Michel has a recurring deposit account in a bank for 2 years at 6% per annum simple interest. If he gets $1200 as interest at the time of maturity find –
(i) the monthly instalment
(ii) the amount of maturity
Ans.) It is given that SI = $1200, R = 6% p.a. and n = 24 months
(i) Let the monthly installment be $P.
1 1 R
Then, SI = $ [P X ------- n (n + 1) X ------- X -------]
2 12 100
1 1 6
=> [P X ------- 24 (24 + 1) X ------- X ------] = 1200
2 12 100
1 1 6
=> [P X ------ 24 X 25 X ------- X -------] = 1200
2 12 100
3P
=> ------- = 1200
2
2
=> P = 1200 X ------ = $ 800
3
Hence the monthly installment is $ 800
(ii) Amount of maturity = $ (800 X 24) + Interest
= $ 19200 + $ 1200 = $ 20400 (Ans.)
Example.3) Mr. Stephen opened a recurring deposit account in a bank and deposited $1000 per month
1
for 1 ----- years. If he received $18684 at the time of maturity, find
2
the rate of interest per annum.
Ans.) Here P = $1000, n = 18 months
As per the given condition, total money deposited = $(1000 X 18)
= $18000
Total money received = $22000
SI = $(18684 – 18000) = $4000
1 1 R
Then, SI = $ [P X ------- n (n + 1) X ------- X -------]
2 12 100
1 1 R
Or, 684 = [1000 X ------ X 18 X (18 + 1) X ------ X ------]
2 12 100
1
Or, 684 = (5 X 18 X 19 X ------ X R)
12
684 X 12
Or, R = --------------- = 4.8%
5 X 18 X 19
Hence, the rate of interest is 4.8% per annum. (Ans.)
Example.4) Mr. Ronald deposits a certain sum of money each month in a recurring deposit account of a bank. If the rate of interest is 8% per annum and Ronald gets $8088 from the bank after 3 years, find the value of his monthly installment.
Ans.) Let, the monthly installment be $P
Total money deposited = $36P
Total money received = $8088
Here, n = 36 months,
R = 8% p.a
SI = $(8088 – 36P)
1 1 R
Then, SI = $[P X ------- n (n + 1) X -------- X --------]
2 12 100
1 1 8
Or, (8088 – 36P) = $ [P X ------X 36 X (36 + 1) X ------- X -------]
2 12 100
1 1 8
Or, (8088 – 36P) = $ [P X ------- X 36 X 37 X ------- X -------]
2 12 100
111P
Or, (8088 – 36P) = $ ---------
25
Or, 111P = $[25 X (8088 – 36P)]
Or, 111P = $(202200 – 900P)
Or, 111P + 900P = $202200
Or, 1011P = $202200
202200
Or, P = $ ----------- = $200
1011
Hence the value of each monthly installment is $200. (Ans.)
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