Q. 165.0( 3 Votes )

Find the number of terms of the A.P. 64, 60, 56, ... so that their sum is 544. Explain the double answer.

Answer :

AP = 64, 60, 56, …


Here, a = 64, d = 60 – 64 = -4






1088 = n[132 – 4n]


4n2 – 132n + 1088 = 0


n2 – 33n + 272= 0


n2 – 16n - 17n + 272 = 0


n(n – 16) - 17(n – 16) = 0


(n – 16)(n – 17) = 0


n – 16 = 0 or n – 17 = 0


n = 16 or n = 17


If n = 16, a = 64 and d = -4


a16 = 64 + (16 – 1)(-4)


a16 = 64 + 15 × -4


a16 = 64 – 60


a16 = 4


and If n = 17, a = 64 and d = -4


a17 = 64 + (17 – 1)(-4)


a17 = 64 + 16 × -4


a17 = 64 – 64


a17 = 0


Now, we will check at which term the sum of the AP is 544.




S16 = 8[68]


S16 = 544


and


S17 = 17 × 32


S17 = 544


So, the terms may be either 17 or 16 both holds true.


We get a double answer because the 17th term is zero and when we add this in the sum, the sum remains the same.


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