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Ex 9.2, 17 - A man starts repaying a loan as first instalment - Arithmetic Progression (AP): Statement

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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Ex9.2,17 A man starts repaying a loan as first instalment of Rs.100. If he increases the instalment by Rs 5 every month, what amount he will pay in the 30th instalment? First instalment = 100. Every month instalment increases by Rs 5 ∴ Second instalment = 100 + 5 = 105 Third instalment = 105 + 5 = 110 So, the instalments are 100, 105, 110, … The instalments is in A.P as difference between consecutive terms is constant. 100, 105, 110, … Here, first term = a = 100 & common difference = 105 – 100 = 5 We need to find 30th instalment To find it, we use the formula an = a + (n – 1)d where an = nth term , n = number of terms, a = first term , d = common difference Here, an = 30th instalment , n = 30 , a = 100, d = 5 a30 = 100 + (30 – 1)5 = 100 + (29) 5 = 100 + 145 = 245 Hence, the 30th instalment is Rs 245

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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