The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
\(S_{16} = 20, 76\)
Step 1: Express the given terms in terms of \(a\) and \(d\).
Let the first term be \(a\) and common difference be \(d\).
Third term: \(a_3 = a + 2d\)
Seventh term: \(a_7 = a + 6d\)
Given:
\[a_3 + a_7 = 6, \quad a_3 \cdot a_7 = 8\]
Step 2: Use the sum condition.
\[(a + 2d) + (a + 6d) = 6\]
\[2a + 8d = 6\]
Divide by 2:
\[a + 4d = 3 \quad ...(1)\]
Step 3: Use the product condition.
\[(a + 2d)(a + 6d) = 8\]
Notice that \(a + 2d\) and \(a + 6d\) are symmetric around \(a + 4d\). Let:
\[x = a + 4d\]
Then:
\[a + 2d = x - 2d, \quad a + 6d = x + 2d\]
From (1), we already have \(x = 3\).
Now the product condition becomes:
\[(x - 2d)(x + 2d) = 8\]
\[x^2 - 4d^2 = 8\]
Substitute \(x = 3\):
\[9 - 4d^2 = 8\]
\[4d^2 = 1 \Rightarrow d^2 = \dfrac{1}{4}\]
So:
\[d = \pm \dfrac{1}{2}\]
Step 4: Find \(a\) for each value of \(d\).
From (1): \(a + 4d = 3\).
Case I: \(d = \dfrac{1}{2}\)
\[a + 4 \cdot \dfrac{1}{2} = 3 \Rightarrow a + 2 = 3 \Rightarrow a = 1\]
Case II: \(d = -\dfrac{1}{2}\)
\[a + 4(-\dfrac{1}{2}) = 3 \Rightarrow a - 2 = 3 \Rightarrow a = 5\]
So two possible APs:
• AP1: \(a = 1, d = \dfrac{1}{2}\)
• AP2: \(a = 5, d = -\dfrac{1}{2}\)
Step 5: Find \(S_{16}\) for each AP.
Formula: \[S_n = \dfrac{n}{2}[2a + (n - 1)d]\]
Here \(n = 16\).
For AP1: \(a = 1, d = \dfrac{1}{2}\)
\[S_{16} = \dfrac{16}{2}\big[2 \cdot 1 + 15 \cdot \dfrac{1}{2}\big]\]
\[S_{16} = 8\big[2 + \dfrac{15}{2}\big] = 8 \cdot \dfrac{4 + 15}{2} = 8 \cdot \dfrac{19}{2} = 4 \cdot 19 = 76\]
For AP2: \(a = 5, d = -\dfrac{1}{2}\)
\[S_{16} = \dfrac{16}{2}\big[2 \cdot 5 + 15 \cdot (-\dfrac{1}{2})\big]\]
\[S_{16} = 8\big[10 - \dfrac{15}{2}\big] = 8 \cdot \dfrac{20 - 15}{2} = 8 \cdot \dfrac{5}{2} = 4 \cdot 5 = 20\]
Conclusion: There are two possible APs satisfying the given conditions, so the sum of the first sixteen terms can be \(S_{16} = 20\) or \(S_{16} = 76\).