Question

Mr. X arranges to pay off a debt of Rs.9600 in 48 installments which form an arithmetical series. When 40 of these installments are paid, Mr. X becomes insolvent and his creditor finds that Rs.2400 still remains unpaid.Find the value of each of the first three installments of Mr. X. Ignore interest.



Please give correct answer.​

Answer 1

Answer:

Correct option is

C

Rs. 51

Answer 2

Answer:

1st instalment =82.50

2nd instalment = 87.50

3rd instalment = 92.50

Step-by-step explanation:

let the first installment be a and common difference of AP be d

given,

9600=sum of 48 terms

Sn=n/2[2a+(n-1) d]

9600=24*2a+47d

2a+47d=400.........1

after 40 installment rupees 2400 is still remains unpaid hence 2400 is unpaid and 7200 is paid

now, case 2:

7200=20*2a+39d

360=2a+39d

2a+39d=360.......2

subtracting equation two from one we get

8d=40

d=5

putting the value of d in equation 1 we get

2a+47*5=400

2a=165

a=82.50

therefore

t1 = 82.50

t2 = 87.50

t3 = 92.50