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3 Marks4. A man arranges to pay off a debt of Rs 7200 in annual installments which form anArithmetic progression. When 15 of these installments have been paid, he dies leavingone-third of his debt unpaid. Find the value of the first installment? |
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Answer» Answer: According to the question, Tthe total amount of debit to be paid in 40 installment = Rs 3600 After 30 installment one-third of his debit is left unpaid. This means that he paid TWO third of the payment. So, The amount he paid in 30 installments = 3 2
(3600) = 2(1200) = 2400 LET us take the first installment as a and COMMON difference as d. So, using the formula for the sum of n terms of an A.p, s n
= 2 n
[2a+(n−1)d] Let us find a and d, for 30 installments. s 30
= 2 30
[2a+(30−1)d] 2400=15[2a+(29)d] 15 2400
=2a+29d 160=2a+29d a= 2 160−29d
.....(1) Similarly, we find a and d for 40 installment. s 40
= 2 40
[2a+(40−1)d] 3600=20(2a+(39)d) 20 3600
=2a+39d a= 2 180−39d
subtracting (1) from (2), we get a−a=( 2 180−39d
)−( 2 160−29d
) 0= 2 180−39d−160+29d
.....(2) 0=20−10d Further solving for d 10d = 20 d= 10 20
d = 2 the value subtracting the value of d in (1), we get, a= 2 160−29(2)
= 2 160−58
= 2 102
= Rs 51 Step-by-step explanation: You should not cheat in the examination. |
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