How many terms of the A.P. 24, 21, 18, … must be taken so that the sum is 78 ? Explain the double answer.

Let numbers of terms be n.
Given, a = 24, d = 21 - 24 = -3 and S_n = 78.

By formula S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ 78 = $\dfrac{n}{2}[2 \times 24 + (n - 1)(-3)]$
⇒ 78 × 2 = n[48 - 3n + 3]
⇒ 156 = n[51 - 3n]
⇒ 156 = 51n - 3n²
⇒ 3n² - 51n + 156 = 0
⇒ 3n² - 12n - 39n + 156 = 0
⇒ 3n(n - 4) - 39(n - 4) = 0
⇒ (3n - 39)(n - 4) = 0
⇒ 3n - 39 = 0 or n - 4 = 0
⇒ 3n = 39 or n = 4
⇒ n = 13 or n = 4.

Let's check sum of 4 terms
⇒ a₄ = a₃ + d
⇒ a₄ = 18 + (-3) = 15

∴ S₄ = 24 + 21 + 18 + 15 = 78.

By formula, a_n = a + (n - 1)d
⇒ a₅ = 24 + (5 - 1)(-3)
⇒ a₅ = 24 + 4(-3)
⇒ a₅ = 24 - 12 = 12.

a₆ = a₅ + d = 12 + (-3) = 9
a₇ = a₆ + d = 9 + (-3) = 6
a₈ = a₇ + d = 6 + (-3) = 3
a₉ = a₈ + d = 3 + (-3) = 0
a₁₀ = a₉ + d = 0 + (-3) = -3
a₁₁ = a₁₀ + d = -3 + (-3) = -6
a₁₂ = a₁₁ + d = -6 + (-3) = -9
a₁₃ = a₁₂ + d = -9 + (-3) = -12.

Taking sum of from 5th term to 13th,

12 + 9 + 6 + 3 + 0 - 3 - 6 - 9 - 12 = 0

Since, the sum from 5tn term to 13th is zero hence, it will not add a value.

∴ n = 13 and 4.

Hence, the number of terms that can be taken for sum to be 78 can be 4 or 13.