Question

(i) Mr. Gulati has a Recurring Deposit Account of Rs. 300 per month. If the rate of interest is 12% and the maturity value of this account is Rs. 8,100; find the time (in years) of this Recurring Deposit Account.
(ii) Mr. Bajaj needs Rs. 30,000 after 2 years. What least money (in multiple of Rs. 5) must be deposit every month in a recurring deposit account to get required money after 2 years, the rate of interest being 8% p.a.?

Answer 1

P = Rs 300    r = 12%  per annum  =  1% per month
Sum on maturity= Rs 8,100.          let n be number of months after which the amount become mature.

the first installment earns interest for n months,  the second one for n-1 months, and the last installment earns interest for 1 month.    The sum is a geometric series with ratio 1+r/100.

Sum = P (1 + r/100)^n + P (1+r/100)^n-1 + P(1+r/100)^n-2+......+P(1+r/100)^1
8,100 = 300 [ 1.01^n + 1.01^n-1  + 1.01^n-2 +.... + 1.01^2 + 1.01 ]
 27 = 1.01 [ 1.01^(n+1) - 1 ] / [ 1.01 - 1]
     
27 = 101 [ 1.01^(n+1)  -  1 ]
       1.01^(n+1)  = 1 +  27/101 = 128/101
        (n+1) Log 1.01 =  Log 128/101
   n+ 1    =  [ log 128/101  ]  / Lo g 1.01 = 23.81

   n = 22.81  months      or,    1.984 years
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sum 30,000            number of installments = 24  in 2 years
     r = 8% per annum =  8/12%  per month 
     r/100 =  0.006667  per month

we use the formula  derived using geometric series here:

30, 000 = P * 1.006667^(n+1)  -  1)  /  (1.006667  -  1)

30, 000 =  P * (1.006667^25 - 1 ) / 0.006667

P = Rs 1, 106.76

The least monthly deposit amount in the RD will be :  Rs 1, 110.  in multiples of 5.  which will become : Rs 30,087.75  in 2 years.