Question 2 - Miscellaneous - 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
Misc 15 Important
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Misc 17 Important
Misc 18 Important
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Miscellaneous
Last updated at April 16, 2024 by Teachoo
Question 2 If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers. Let the three numbers in A.P. be a d, a, a + d. Sum of the three numbers is 24 (a d) + (a) + (a + d) = 24 a + a + a + d d = 24 3a + 0 = 24 3a = 24 a = 24/3 a = 8 Also, it is given that product of three number is 440 i.e. (a d) (a) (a + d) = 440 (a d)(a + d)a = 440 (a2 d2)a = 440 Putting a = 8 (82 d2)8 = 440 82 d2 = 440/8 64 d2 = 55 64 55 = d2 9 = d2 d2 = 9 d = 9 d = 3 So, d = 3 , d = 3 When d = 3 & a = 8 The numbers are a d = 8 3 = 5 a = 8 a + d = 8 + 3 = 11 Hence the numbers are 5, 8, 11 When d = -3 & a = 8 The number are a d = 8 (-3) = 8 + 3 = 11 a = 8 a + d = 8 + (-3) = 8 3 = 5 Hence the numbers are 11, 8, & 5 Hence the numbers are 5, 8, 11 for a = 8, d = 3 and 11, 8, & 5 for a = 8, d = -3