ICSE Class 10 Answered
How many terms of the A P 17,15,13,11..... must be added to get the sum 72?
Asked by suneelmsc | 29 Jun, 2020, 16:59: PM
In this AP, we are given a = 17, d = -2, Sn = 72 and we have to find n
Sum of first n terms = n/2[2a+(n-1)d]
Sum is given as 72
72 = n/2[2 x 17 + (n - 1)(-2)]
144 = n[34-2n+2]
144 = n[36-2n]
72 = n(18 - n)
n2 - 18n + 72 = 0
n2 - 12n - 6n + 72 = 0
n(n - 12) - 6(n - 12) = 0
(n - 12)(n - 6) = 0
=> n = 6, 12
Both values of n, being positive integers are valid. We get double answer because sum of 7th to 12th terms is zero, as some terms are positive and some are negative
Both values of n, being positive integers are valid. We get double answer because sum of 7th to 12th terms is zero, as some terms are positive and some are negative
Answered by Renu Varma | 29 Jun, 2020, 20:05: PM
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