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he save in the llist mou7. A man arranges to pay off a debt of6000by 40 monthly instalmentswhich form an arithmetic series. When 30 of the instalments are paid,he dies leaving one-third of the debt unpaid. Find the value of the firstinstalment. |
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Answer» Let the first installments be aGiven that the installments are in AP.Let the common difference be dHence the installments are: a, a + d, a + 2d ………..40th termRecall the nth term of AP, tn = a + (n – 1)d⇒ t40 = a + (40 – 1)d = a + 39dSum of n term of AP, Given total debt = Rs 3600⇒ 20[2a + 39d] = 36002a + 39d = 180 → (1)It is also given that 30 installments are paid.Unpaid amount = one-third of Rs 3600 = Rs 1200Therefore total payment till 30 installments = 3600 – 1200 = Rs 2400Now, sum of 30 terms, = 15(2a + 29d)⇒ 15(2a+29d) = 24002a +29d = 160 → (2)Subtract (2) and (1), we get2a + 39d = 1802a +29d = 160-------------------- 10d = 20∴ d = 2Put d = 2 in equation (1)2a + 29(2) = 180⇒ 2a + 58 = 180⇒ 2a = 180 – 58 = 122∴ a = 618th installment, t8 = a + 7d= 61 + 7(2)= 61 + 14 = 75 |
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