Jaspal Singh repays a loan of Rs 118000 by monthly instalments starting with Rs 1000 and increasing by Rs 100 every month. What will he pay in the 30th instalment? What amount is still unpaid after the 30th instalment?
30th instalment: Rs 3900; Unpaid after 30th: Rs 44,500
Step 1: Identify the pattern of payments
Jaspal Singh starts by paying Rs 1000 in the first month. Each month he pays Rs 100 more than the previous month. So, the instalments are:
1000, 1100, 1200, 1300, …
This is an arithmetic progression (AP) where:
Step 2: Find the 30th instalment
The formula for the \(n^{th}\) term of an AP is:
\(a_n = a + (n-1)\times d\)
For the 30th instalment (\(n = 30\)):
\(a_{30} = 1000 + (30-1)\times100 = 1000 + 2900 = 3900\)
So, the 30th instalment is Rs 3900.
Step 3: Find the total paid in 30 months
The formula for the sum of first \(n\) terms of an AP is:
\(S_n = \dfrac{n}{2}\,[2a + (n-1)d]\)
For \(n = 30\):
\(S_{30} = \dfrac{30}{2}[2(1000) + (30-1)(100)]\)
\(= 15[2000 + 2900]\)
\(= 15 \times 4900\)
\(= 73500\)
So, the total paid in 30 months = Rs 73,500.
Step 4: Find the unpaid amount
Total loan = Rs 118000
Amount paid in 30 months = Rs 73500
Unpaid = Rs 118000 – Rs 73500 = Rs 44,500
Final Answer:
30th instalment = Rs 3900
Unpaid after 30 instalments = Rs 44,500