A man arranges to pay off a debt of ₹ 3600 in 40 monthly installments which form an AP. When 30 of the installments are paid, he dies leaving one-third of the debt unpaid. Find the value of the first installment.
Answers
Solution :
Monthly installment from in AP.
Let the amount of first installment be a and d be the common difference between the installments which from an AP.
Total amount, S₄₀ = 3600
Using S_n = n/2[2a + ( n - 1)d]
Substituting the values in the above formula, we get,
S₄₀ = 40/2 [2a + (40 - 1)d]
3600 = 20[2a + 39d]
2a + 39d = 3600/20
2a + 39d = 180 ..... (i)
Amount paid in 30 installments ,
= 3600 - 1/2(3600)
= 3600 - 1200
= 2400
Therefore, Amount paid in 30 installments = ₹ 2400.
So, S₃₀ = 30/2[2a + (30-1)d]
2400 = 15[2a + 29d]
2a + 29d = 2400/15
2a + 29d = 160 ..... (ii)
Subtracting (ii) from(i), we get
10d = 20
d = 20/10
d = 2.
•°• From (i), 2a + 39 × 2 = 180
2a = 180 - 78
2a = 102
a = 102/2
a = 51.
Therefore, the value of first installment = ₹ 51.
Step-by-step explanation:
Given :➡️
Total amount of Debt , S40 = 3600
Number of annual installments, n = 40
He paid 30 installment and he dies leaving 1/3 of the debt unpaid.
Unpaid amount = ⅓ × 3600 = 1200
Total payment he paid in 30 installment , S30 = 3600 - 1200 = 2400
S30 = 2400
By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]
For 30 installments :
S30 = 30/2 [2a + (30 - 1)d]
2400 = 15 [2a + 29d]
2400/15 = [2a + 29d]
160 = 2a + 29d
2a = 160 - 29d
2a + 29d = 160 ……..(1)