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.
Step-by-Step Solution
Key Concept: Use the nth‑term formula $T_n = a + (n-1)d$ to express the 3rd and 7th terms in terms of the first term $a$ and common difference $d$. Form two equations from the given sum and product, solve for $a$ and $d$, and then use the sum formula $S_n = \frac{n}{2}[2a+(n-1)d]$ for $n=16$.
1. Let the first term be $a$ and the common difference be $d$.
2. \(T_3 = a + 2d\) and \(T_7 = a + 6d\).
3. Given:
\[T_3 + T_7 = 6 \quad\Rightarrow\quad (a+2d)+(a+6d)=2a+8d=6 \;\Rightarrow\; a+4d=3 \tag{1}\]
\[T_3\times T_7 = 8 \quad\Rightarrow\quad (a+2d)(a+6d)=8 \tag{2}\]
4. Expand (2):
\[(a+2d)(a+6d)=a^2+8ad+12d^2=8\]
5. From (1), $a = 3-4d$. Substitute in the expanded product:
\[ (3-4d)^2 + 8(3-4d)d + 12d^2 = 8 \]
Simplify:
\[9 -24d +16d^2 + 24d -32d^2 +12d^2 = 8 \]
\[9 -4d^2 = 8 \]
\[-4d^2 = -1 \]
\[d^2 = \frac{1}{4} \]
Hence \(d = \frac{1}{2}\) or \(d = -\frac{1}{2}\).
6. Find the corresponding $a$ using (1):
- If \(d = \frac{1}{2}\), then \(a = 3 - 4\left(\frac{1}{2}\right) = 1\).
- If \(d = -\frac{1}{2}\), then \(a = 3 - 4\left(-\frac{1}{2}\right) = 5\).
7. Sum of first $n$ terms of an AP:
\[S_n = \frac{n}{2}\bigl[2a + (n-1)d\bigr]\]
For $n = 16$:
\[S_{16} = 8\bigl[2a + 15d\bigr]\]
- For \(a = 1, d = \frac{1}{2}\):
\[S_{16}=8\bigl[2(1)+15\left(\frac{1}{2}\right)\bigr]=8\bigl[2+7.5\bigr]=8\times9.5=76\]
- For \(a = 5, d = -\frac{1}{2}\):
\[S_{16}=8\bigl[2(5)+15\left(-\frac{1}{2}\right)\bigr]=8\bigl[10-7.5\bigr]=8\times2.5=20\]
8. Therefore the sum of the first sixteen terms can be either $\boxed{76}$ (when $a=1, d=\frac12$) or $\boxed{20}$ (when $a=5, d=-\frac12$). Both APs satisfy the given conditions.
Correct Answer: The sum of the first sixteen terms is 76 for the AP with $a=1$, $d=\frac12$, and 20 for the AP with $a=5$, $d=-\frac12$.