Sheela has a recurring deposit account in a bank of 2000 per month at the rate of 10% per annum. If she gets 83100 at the time of maturity, find the total time (in years) for which the account was held.
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P =2000
R=10%
M.V.=83100
LET THE NO. OF MONTHS BE n
I =P×n(n+1)÷12×2×rate
I =(25n2+25n)÷3
M.V.=P×n+I
83100=2000×n+(25n2+25n)÷3
83100×3=6000n+25n2+25n
249300=6025n +25n2
249300=25(241n+n2)
249300÷25=241n+n2
9972=241n+n2
0=n2+241n-9972
0=n2+(277-36)n-9972
0=n2+277n -36n-9972
0=n(n+277)-36(n+277)
0=(n-36)(n+277)
0=n-36
n=36
THE ACCOUNT WAS HELD FOR 3 YEARS
P.S. -277 is neglected because time can not be negative.
R=10%
M.V.=83100
LET THE NO. OF MONTHS BE n
I =P×n(n+1)÷12×2×rate
I =(25n2+25n)÷3
M.V.=P×n+I
83100=2000×n+(25n2+25n)÷3
83100×3=6000n+25n2+25n
249300=6025n +25n2
249300=25(241n+n2)
249300÷25=241n+n2
9972=241n+n2
0=n2+241n-9972
0=n2+(277-36)n-9972
0=n2+277n -36n-9972
0=n(n+277)-36(n+277)
0=(n-36)(n+277)
0=n-36
n=36
THE ACCOUNT WAS HELD FOR 3 YEARS
P.S. -277 is neglected because time can not be negative.
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