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A sum of\$ 5000 is invested a 6 percent per annum simple interest. Calculate the interest at the end of year and I show that they form an AP. Also,find the interest at the end of 25th year.
Asked by kipgenhaolenthang | 29 Apr, 2020, 07:46: AM
Here P = \$ 5000, R = 6%, N=1, 2, 3, 4, ...
At the end of 1st year
I = P x N x R/100 = (5000 x 6 x 1)/100 = \$ 300
At the end of 2nd year
I = P x N x R/100 = (5000 x 6 x 2)/100 =\$ 600
At the end of 3rd year
I = P x N x R/100 = (5000 x 6 x 3)/100 = \$ 900
At the end of 4th year
I = P x N x R/100 = (5000 x 6 x 4)/100 = \$ 1200
and so on...
300, 600, 900, 1200, ... forms a series
Now, 600-300=300, 900-600=300, 1200-900=300
i.e. the diference is same for any two successive terms.
Therefore, 300, 600, 900, 1200, ... forms an AP.
We know that an = a+(n-1)d
Therefore,
a25 = a+(25-1)d
= 300 + 24 x 300
= 300 + 7200
= 7500
Hence, the interest at the end of 25th year is \$ 7500.
Answered by Renu Varma | 29 Apr, 2020, 01:52: PM

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