Arithmetic Progression Question (Class 10). Solve this.
A man pays off a debt of Rs 3600 by 40 annual 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.
Please prove step by step answer. Direct answers will be reported.
Answers
Answer.
Let the first installments be a Given that the installments are in AP. Let the common difference be d Hence the installments are: a, a + d, a + 2d ………..40th term Recall the nth term of AP, tn = a + (n – 1)d ⇒ t40 = a + (40 – 1)d = a + 39d Sum of n term of AP, Given total debt = Rs 3600 ⇒ 20[2a + 39d] = 3600 2a + 39d = 180 → (1) It is also given that 30 installments are paid. Unpaid amount = one-third of Rs 3600 = Rs 1200 Therefore total payment till 30 installments = 3600 – 1200 = Rs 2400 Now, sum of 30 terms, = 15(2a + 29d) ⇒ 15(2a+29d) = 2400 2a +29d = 160 → (2) Subtract (2) and (1), we get 2a + 39d = 180 2a +29d = 160 -------------------- 10d = 20 ∴ d = 2 Put d = 2 in equation (1) 2a + 29(2) = 180 ⇒ 2a + 58 = 180 ⇒ 2a = 180 – 58 = 122 ∴ a = 61 8th installment, t8 = a + 7d = 61 + 7(2) = 61 + 14 = 75