Arithmetic Progressions | Exercise 5.4

Question 2

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.

Solution

Understand the Question

Let $a$ be the first term and $d$ be the common difference of the AP.
The $n^{\text{th}}$ $n^{th}$ term of an AP is given by $a_n = a + (n - 1)d$ $a_{n} = a + (n - 1) d$ , so:
- $a_3 = a + 2d$
- $a_7 = a + 6d$
We are given two conditions:
1. Sum: $a_3 + a_7 = 6$
2. Product: $a_3 \cdot a_7 = 8$
Solving these equations gives two possible values for $d$ ( $+\dfrac{1}{2}$ and $-\dfrac{1}{2}$ ) and corresponding values for $a$ .
Finally, we calculate the sum of the first $16$ terms using the formula $S_n = \dfrac{n}{2}[2a + (n-1)d]$ for both cases.

Step 1 · Formulate Equations for $a$ and $d$

Using the formula $a_n = a + (n-1)d$ : $a_3 = a + 2d, \quad a_7 = a + 6d$

From the given sum:

\begin{aligned} a_3 + a_7 &= 6 \\ (a + 2d) + (a + 6d) &= 6 \\ 2a + 8d &= 6 \\ a + 4d &= 3 \\ a &= 3 - 4d \quad \dots (1) \end{aligned}

From the given product:

\begin{aligned} (a_3)(a_7) &= 8 \\ (a + 2d)(a + 6d) &= 8 \quad \dots (2) \end{aligned}

Step 2 · Solve for the Common Difference $d$

Substitute $a = 3 - 4d$ from equation $(1)$ into equation $(2)$ :

\begin{aligned} ((3 - 4d) + 2d)((3 - 4d) + 6d) &= 8 \\ (3 - 2d)(3 + 2d) &= 8 \\ 3^2 - (2d)^2 &= 8 \\ 9 - 4d^2 &= 8 \\ 4d^2 &= 9 - 8 \\ 4d^2 &= 1 \\ d^2 &= \dfrac{1}{4} \\[0.6em] d &= \pm \dfrac{1}{2} \end{aligned}

Step 3 · Find First Term $a$ for Each Case

Case 1: When $d = \dfrac{1}{2}$

\begin{aligned} a &= 3 - 4\left(\dfrac{1}{2}\right) \\[0.6em] &= 3 - 2 \\[0.6em] &= 1 \end{aligned}

Case 2: When $d = -\dfrac{1}{2}$

\begin{aligned} a &= 3 - 4\left(-\dfrac{1}{2}\right) \\[0.6em] &= 3 + 2 \\[0.6em] &= 5 \end{aligned}

Step 4 · Calculate Sum of First 16 Terms ( $S_{16}$ )

Using the sum formula $S_n = \dfrac{n}{2}[2a + (n-1)d]$ :

Case 1: For $a = 1$ and $d = \dfrac{1}{2}$

\begin{aligned} S_{16} &= \dfrac{16}{2}\left[2(1) + (16-1)\left(\dfrac{1}{2}\right)\right] \\[0.6em] &= 8\left[2 + 15\left(\dfrac{1}{2}\right)\right] \\[0.6em] &= 8\left[2 + \dfrac{15}{2}\right] \\[0.6em] &= 8\left[\dfrac{4 + 15}{2}\right] \\[0.6em] &= 8\left[\dfrac{19}{2}\right] \\[0.6em] &= 4 \times 19 \\[0.6em] &= 76 \end{aligned}

Case 2: For $a = 5$ and $d = -\dfrac{1}{2}$

\begin{aligned} S_{16} &= \dfrac{16}{2}\left[2(5) + (16-1)\left(-\dfrac{1}{2}\right)\right] \\[0.6em] &= 8\left[10 + 15\left(-\dfrac{1}{2}\right)\right] \\[0.6em] &= 8\left[10 - \dfrac{15}{2}\right] \\[0.6em] &= 8\left[\dfrac{20 - 15}{2}\right] \\[0.6em] &= 8\left[\dfrac{5}{2}\right] \\[0.6em] &= 4 \times 5 \\[0.6em] &= 20 \end{aligned}

Answer

$76$ or $20$

Common Mistakes

Ignoring the Negative Root: Taking only $d = \dfrac{1}{2}$ and forgetting $d = -\dfrac{1}{2}$ from $d^2 = \dfrac{1}{4}$ , which leads to missing the second valid sum $S_{16} = 20$ .
Sign Errors in Multiplication: Making mistakes with signs when evaluating $a = 3 - 4\left(-\dfrac{1}{2}\right) = 3 + 2 = 5$ or in the $S_{16}$ calculation.

More questions in Exercise 5.4