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Example 9 - A sum of Rs 1000 is invested at 8 simple interest - Examples

  1. Chapter 5 Class 10 Arithmetic Progressions
  2. Serial order wise
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Example 9 A sum of Rs 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact. We know that Interest = (๐‘ƒ ร— ๐‘… ร— ๐‘‡)/100 Here P = Principal , R = Rate , T = Time Given ,P = 1000 , R = 8 % & T = T years Interest at the end of 1st year = (1000 ร— 8 ร— 1)/100 = Rs 80 Interest at the end of 2rd year = (1000 ร— 8 ร— 2)/100 = Rs 160 Interest at the end of 3rd year = (1000 ร— 8 ร— 3)/100 = Rs 240 Therefore, interest at the end of year (1st , 2nd,3rd , โ€ฆ.)forms a series 80, 160 , 240, โ€ฆโ€ฆโ€ฆ. Difference between 2nd and 1st term = 160 โ€“ 80 = 80 Difference between 3rd and 2nd term = 240 โ€“ 160 = 80 Since difference is same Hence it is an AP We have to find interest at the end of 30 years . We know an = a + (n โ€“ 1) d Here , a = 80 , d = 80 , n = 30 Putting these values in formula an = a + (n โ€“ 1)d a30 = 80 + (30 โ€“ 1) 80 = 80 + 29 ร—80 = 80 + 2320 = 2400 So, the interest at the end of 30 years = Rs 2400

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Davneet Singh
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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