Question

A person borrows Rs.8,000 at 2.76% simple interest per annum. The principal

and the interest are to be paid in 10 monthly installments. If each installment is

double the preceding one, the value of the first and the last installments are:

Answer 1

Given :-

• A person borrows Rs.8,000 at 2.76% simple interest per annum.
• The principal and the interest are to be paid in 10 monthly installments.
• Each installment is double the preceding one.

To Find :-

• The value of the first and the last installments ?

Solution :-

→ Principal = Rs.8000

→ Time = 10 months. = (10/12) = (5/6) years.

→ Rate = 2.76% per annum.

So,

→ Simple interest = (Principal * Rate * Time)/100

→ SI = (8000 * 2.76 * 5) / (100 * 6)

→ SI = (80 * 2.76 * 5) / 6

→ SI = Rs.184 .

Therefore,

→ Amount to be paid = Principal + Simple interest .

→ Amount = 8000 + 184

→ Amount = Rs.8184.

So, the person has to give Rs.8184 back in 10 installments .

Now, we have given that, Each installment is double of the preceding one.

So,

→ Let first installment = a .

Than,

→ second installment = double of a = 2a

→ Third installment = double of second installment = 2 * 2a = 4a .

Therefore,

→ 10 installments are = a , 2a, 4a, ____________ 2⁹a .

Sum of all installments will be equal to Amount to be paid.

So,

→ a + 2a + 4a + __________ 2⁹a = 8184

→ a(1 + 2 + 4 + ___________ 2⁹) = 8184

{ This is a GP series with first term as 1 and common ratio as 2. So, using sum of n terms of GP = a(r^n - 1)/(r - 1) . }

Putting all values we get,

→ a[ 1(2¹⁰ - 1)/(2 - 1) ] = 8184

→ a[ 2¹⁰ - 1 ] = 8184

→ a(1024 - 1) = 8184

→ 1023a = 8184

→ a = 8.

Hence,

→ First installment = a = 8 (Ans.)

→ Last installment = 2⁹ * a = 512 * 8 = 4096 .(Ans.)

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