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Question
One person borrows ₹ 4,000 and agrees to repay with a total interest of ₹ 500 in 10 instalments. Each instalment being less than the preceding instalment by ₹ 10. What should be the first and the last instalments?
Solution
The instalments are in A. P.
Amount repaid in 10 instalments (S10) = Amount borrowed + total interest
∴ S10 = 4000 + 500 = 4500
Number of instalments (n) = 10
Each instalment is less than the preceding instalment by ₹ 10.
∴ d = – 10
Sn = `"n"/2 [2"a" + ("n" - 1)"d"]`
∴ S10 = `10/2 [2"a" + (10 - 1) (-10)]`
∴ 4500 = 5[2a + 9(–10)]
∴ `4500/5` = 2a – 90
∴ 900 = 2a – 90
∴ 2a = 900 + 90
∴ 2a = 990
∴ a = `990/2`
∴ a = 495
Now, tn = a + (n – 1)d
∴ t10 = 495 + (10 – 1) (–10)
∴ t10 = 495 + 9(–10)
∴ t10 = 495 – 90
∴ t10 = 405
∴ Amount of the first instalment is 495 and that of the last instalment is 405.
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Case Study Push-ups are a fast and effective exercise for building strength. These are helpful in almost all sports including athletics. While the push-up primarily targets the muscles of the chest, arms, and shoulders, support required from other muscles helps in toning up the whole body.
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- Form an A.P representing the number of push-ups per day and hence find the minimum number of days he needs to practice before the day his goal is accomplished?
- Find the total number of push-ups performed by Nitesh up to the day his goal is achieved.
