Question

find the number of terms of A.P. 64, 60, 56...... so that their sum is 544

Answer 1

AP = 64, 60, 56,............
a = 64, d = -4
Sn = n ÷ 2 (2a + (n - 1) d)
544= n ÷ 2 (2×64 + ( n - 1) - 4)
544 = n ÷ 2 ( 128 - 4n + 4)
544 = n ÷ 2 × 4 ( 32 - n +1)
544 = n × 2 (32 - n + 1)
544 ÷ 2 = n (33 - n)
272 = 33n - n2
n2 - 33n + 272 = 0
n2 - 16n - 17n + 272 = 0
n ( n - 16) - 17 ( n - 16) =0
( n - 16) ( n - 17) = 0
n = 16, n = 17
Put the value of n in
Sn = n÷2 (2a + ( n - 1) d)
544 = n ÷ 2 ( 2 × 64 + ( n - 1) - 4)
Put n = 16, 17 one by one and one from 16, 17 which satisfied will be number of terms

Answer 2

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