Find the number of terms of the A.P. 64, 60, 56, ... so that their sum is 544. Explain the double 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 17^th term is zero and when we add this in the sum, the sum remains the same.