Misc 14 - Chapter 8 Class 11 Sequences and Series
Last updated at April 16, 2024 by Teachoo
Miscellaneous
Misc 2
Misc 3
Misc 4 Important
Misc 5
Misc 6
Misc 7 Important
Misc 8
Misc 9
Misc 10 Important
Misc 11 (i) Important
Misc 11 (ii)
Misc 12 Important
Misc 13
Misc 14 Important You are here
Misc 15 Important
Misc 16
Misc 17 Important
Misc 18 Important
Question 1 Deleted for CBSE Board 2025 Exams
Question 2 Deleted for CBSE Board 2025 Exams
Question 3 Important Deleted for CBSE Board 2025 Exams
Question 4 Deleted for CBSE Board 2025 Exams
Question 5 Deleted for CBSE Board 2025 Exams
Question 6 Important Deleted for CBSE Board 2025 Exams
Question 7 Deleted for CBSE Board 2025 Exams
Question 8 Deleted for CBSE Board 2025 Exams
Question 9 Important Deleted for CBSE Board 2025 Exams
Question 10 Deleted for CBSE Board 2025 Exams
Question 11 Important Deleted for CBSE Board 2025 Exams
Question 12 Deleted for CBSE Board 2025 Exams
Question 13 Important Deleted for CBSE Board 2025 Exams
Question 14 Deleted for CBSE Board 2025 Exams
Miscellaneous
Last updated at April 16, 2024 by Teachoo
Misc 14 Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual installment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him? Amount paid to buy scooter = Rs 22000 He pay cash = Rs 4000 Remaining balance = Rs 22000 – Rs 4000 = Rs 18000 Annual instalment = 1000 + "Interest on unpaid amount @10%" Thus , our instalments are 2800 , 2700 , 2600,… Total number of instalments = (𝑅𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 𝑏𝑎𝑙𝑎𝑛𝑐𝑒 𝑙𝑒𝑓𝑡)/(𝐵𝑎𝑙𝑎𝑛𝑐𝑒 𝑐𝑙𝑒𝑎𝑟𝑒𝑑 𝑝𝑒𝑟 𝑖𝑛𝑠𝑡𝑎𝑙𝑚𝑒𝑛𝑡) = 18000/1000 = 18 So, our instalments are 2800 , 2700 , 2600 , … to 18 terms We can observe that this is an AP as difference between consecutive terms is an AP Here first term (a) = 2800 common difference = d = 2700 – 2800 = – 100 number of terms = n = 18 We need to calculate total amount paid in 18 instalments i.e. (2800 + 2700 +2600 + … to 18 terms) We use the formula, Sn = 𝑛/2 (2a + (n – 1)d) Where, Sn = sum of n terms of A.P. n = number of terms a = first term and d = common difference Putting value of n = 18 , a = 2800 & d = –100 S18 = 18/2 [ 2(2800)+(18 – 1) (-100) ] = 9[ 5600 + 17(-100)] = 9[ 5600 – 1700] = 9(3900) = 35100 Hence, amount paid in 18 instalments = Rs 35100 Total amount paid by him = Amount paid earlier + Amount paid in 18 instalments = 4000 + 35100 = Rs 39100