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24Arunrepayshisloan amount of1,18,000 by paying every month,starting with the first installment of .1000. If he increases theinstallment by .100 every month what amount will be paid as thelast installment of loan?. |
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Answer» Loan Amount = Rs 118000 First installment = 1000Second installment = 1100Third installment = 1200 Thus loan installments form an AP with first digit a = 1000, common difference d = 100 Let number of installments are n to repay loan Then,Sn = (2a + (n-1)d)n/2118000*2 = [2*1000 + (n-1)100]n1180*2 = n(20 + (n-1))2360 = 19n + n^2 n^2 + 19n - 2360 = 0n^2 + 59n - 40n - 2360 = 0n(n + 59) - 40(n + 59) = 0n = 40,-59Negative value not possible So total number of installments are 40 Amount paid for last installment= 1000 + (39)*100= Rs 4900 |
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