A man arranges to pay off a debt of Rs. 3600 by 40 annual installments which form an arithmetic series. When 30 of the installments are paid, he dies leaving one third of the debt unpaid, find the value of the first installment.
A
Rs. 53
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B
Rs. 54
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C
Rs. 50
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D
Rs. 51
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Solution
The correct option is DRs. 51 Let the first instalment = Rs. a Common difference between installment = Rs. d. Case 1: A man pays of a debit of Rs. 3600 in 40 annual statements. ∴n=40,S40= Rs. 3600 Sn=n2[2a+(n−1)d] S40=402[2a+(40−1)d] 3600=20[2a+39d] 2a+39d=180 .....(1) Case 2: Man die leaving one third of installment unpaid. Then unpaid amount =13×3600=1200 Hence, debt paid in 30 installment =3600−1200=2400 Heren=30,s30=2400 Sn=n2[2a+(n−1)d] S30=302[2a+(30−1)d] 2400=15[2a+29d] 2a+29d=160 ........(2) Subtracting equation (1) by equation (2), 10d=20 ⇒d=2 Putting the value of d in equation (1), we get ⇒2a+39d=180 ⇒2a+39×2=180 ⇒2a=102 ⇒a=51 Hence, the first term paid by man = Rs. 51.