How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78?
Step-by-Step Solution
Key Concept: Use the formula for the sum of the first \(n\) terms of an arithmetic progression: \(S_n = \frac{n}{2}[2a + (n-1)d]\). This leads to a quadratic equation in \(n\) whose integer solutions give the required number of terms.
1. Identify the first term \(a\) and common difference \(d\):
\[ a = 24, \quad d = 21-24 = -3. \]
2. Write the sum formula for \(n\) terms and set it equal to 78:
\[ S_n = \frac{n}{2}[2a + (n-1)d] = 78. \]
Substituting \(a\) and \(d\):
\[ \frac{n}{2}[2\times24 + (n-1)(-3)] = 78. \]
3. Simplify the expression inside the brackets:
\[ 2\times24 = 48, \quad (n-1)(-3) = -3n + 3. \]
Hence,
\[ \frac{n}{2}[48 - 3n + 3] = \frac{n}{2}[51 - 3n] = 78. \]
4. Multiply both sides by 2 to clear the denominator:
\[ n(51 - 3n) = 156. \]
5. Expand and bring all terms to one side:
\[ 51n - 3n^2 = 156 \]
\[ 3n^2 - 51n + 156 = 0. \]
6. Divide the quadratic by 3 for simplicity:
\[ n^2 - 17n + 52 = 0. \]
7. Factorise the quadratic:
\[ (n - 13)(n - 4) = 0. \]
Hence, \(n = 13\) or \(n = 4\).
8. Verify both values:
- For \(n = 4\):
\[ S_4 = \frac{4}{2}(24 + 15) = 2 \times 39 = 78. \]
- For \(n = 13\):
Last term \(a_{13} = 24 + 12(-3) = -12\).
\[ S_{13} = \frac{13}{2}(24 + (-12)) = \frac{13}{2} \times 12 = 13 \times 6 = 78. \]
Both satisfy the condition.
9. Therefore, the required number of terms can be either 4 or 13.
Correct Answer: 4 terms or 13 terms