Preeti has a recurring deposit account of Rs 1000 per month at 10%per annum. If she gets Rs 5550 as interest at the time of maturity, fund the total time for which the account was held.

Formula is this:SI =Px1/2n(n+1)x 1/12xR /100

### Asked by bejibickey8482.10sdatl | 22nd Apr, 2020, 12:25: PM

Recurring deposit interest I is given by,

I = P×[ n(n+1)/(2×12) ] (r/100) .............. (1)

where P is monthly contribution, P = Rs. 1000

n = number of months,

r = rate of interest = 10%

and I = interest = Rs.5550

By substituting the values in eqn.(1), we get

1000×[ n(n+1)/24 ]×(10/100) = 5550 or n(n+1) = (24/100)×5550 = 1332

Therefore, n^{2} + n - 1332 = (n+37) (n-36) = 0

n = -37 or n= 36

Since, n can't be negative.

Hence, the number of months = 36

### Answered by Renu Varma | 23rd Apr, 2020, 10:12: AM

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