## A sum of Rs 1000 is invested at 8% per annum simple interest. Calculate the Interest at the end of 1,2,3… years. Is the sequence of i

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## Answers ( )

GIVEN:-⟹ P = ₹ 1000

⟹ R = 8%

⟹ T = 1, 2 and 3

TO FIND:-SOLUTION:-S.I = P×R×T/100So,

The interest at the end of the 1st year:

⟹ ₹ 1000×8×1/100

⟹

₹80The interest at the end of the 2nd year:

⟹ ₹ 1000×8×2/100

⟹

₹160The interest at the end of the 3rd year:

⟹ ₹ 1000×8×3/100

⟹

₹240Therefore, the interest(in ₹) at the end of 1st, 2nd and 3rd year years are

80,160,240.It is an AP is the difference between the consecutive terms in the list is 80, that is d = 80 also a = 80.

So, to find the interest at the end of 30 years we shall find a30.

a30 = a+(30-1)d

a30 = 80+29×80

a30 = 2400

So, the interest at the end of 30 years will be

₹2400.Given,

principal=1000

rate of interest=8%

number of years=1 year

SI=Pnr/100=1000x1x8/100=rs.80

when n=2 years,SI=1000x2x8/100=rs.160

when n=3years,SI=1000x3x8/100=rs.240

the sequence is

80,160,240,……….

this is in AP as a,first term=80,d,common difference=80

interest at the end of 30 years,t30=a+(n-1)d

=80+(30-1)80

=80+29(80)

=80+2320

=2400

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