Here, a = 15, d = 12 – 15 = -3 and Sn = 15
We know that,
⇒ 30 = n[33 – 3n ]
⇒ 3n2 – 33n + 30 = 0
⇒ 3n2 – 30n – 3n + 30 = 0
⇒ 3n(n – 10) -3(n – 10) = 0
⇒ (n – 10)(3n – 3) = 0
⇒ n – 10 = 0 or 3n – 3 = 0
⇒ n = 10 or n = 1
The number of terms can be 1 or 10.
Here, the common difference is negative.
∴ The AP starts from a positive term, and its terms are decreasing.
∴ All the terms after 6th term are negative.
We get a double answer because these positive terms from 2nd to 5th term when added to negative terms from 7th to 10th term, they cancel out each other and the sum remains same.