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 | 29th Apr, 2020, 07:46: AM

Expert Answer:

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 | 29th Apr, 2020, 01:52: PM

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