Deepak repays his total loan of Rs 1 ,18,000 by paying every month sta

1.	Deepak repays his total loan of Rs 1 ,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month. what amount will be paid as the last instalment of loan?(A) Rs 4900(B) Rs 5400(C) Rs 3500(D) Rs 4500
Answer» The correct option is: (A) Rs 4900 Explanation: 1^st instalment = Rs 1000 2^nd instalment = Rs1000 + Rs 100 = Rs 1100 3^rd instalment = Rs 1100 + Rs 100 = Rs 1200 and so on Let number of instalments = n .'. 1000 + 1100 + 1200 + ... up to n terms = 118000 => n/2[2 x 1000 + (n - 1) 100] = 118000 => 100n² + 1900n - 236000 = 0 => n²+ 19n - 2360 = 0 => (n + 59)(n - 40) = 0 .'. Total no. of instalments = 40 Now, last instalment = 40^th instalment . ^.. a₄₀ = a + 39d = Rs 4900

Deepak repays his total loan of Rs 1 ,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month. what amount will be paid as the last instalment of loan?(A) Rs 4900(B) Rs 5400(C) Rs 3500(D) Rs 4500

Answer»

The correct option is: (A) Rs 4900

Explanation:

1^st instalment = Rs 1000

2^nd instalment = Rs1000 + Rs 100 = Rs 1100

3^rd instalment = Rs 1100 + Rs 100 = Rs 1200 and so on

Let number of instalments = n

.'. 1000 + 1100 + 1200 + ... up to n terms = 118000

=> n/2[2 x 1000 + (n - 1) 100] = 118000

=> 100n² + 1900n - 236000 = 0

=> n²+ 19n - 2360 = 0

=> (n + 59)(n - 40) = 0

.'. Total no. of instalments = 40

Now, last instalment = 40^th instalment

. ^.. a₄₀ = a + 39d = Rs 4900