The sum of first and eighth terms of an A.P. is 32 and their product is 60. Find the first term and common difference of the A.P. Hence, also find the sum of its first 20 terms.
Generated by claude-sonnet-4-6 · 2026-06-15 10:34 · grounding rag
Model Answer
Let the first term = $a$ and common difference = $d$.
8th term: $a_8 = a + 7d$
Given:
$$a + a_8 = 32 \Rightarrow a + (a + 7d) = 32 \Rightarrow 2a + 7d = 32 \quad ...(1)$$
$$a \cdot a_8 = 60 \Rightarrow a(a + 7d) = 60 \quad ...(2)$$
From (1): $a + (a + 7d) = 32$, so let $a = x$ and $a + 7d = y$, where $x + y = 32$ and $xy = 60$.
These are roots of: $t^2 - 32t + 60 = 0$
$$t = \frac{32 \pm \sqrt{1024 - 240}}{2} = \frac{32 \pm \sqrt{784}}{2} = \frac{32 \pm 28}{2}$$
So $t = 30$ or $t = 2$.
Case 1: $a = 30$, $a + 7d = 2 \Rightarrow d = -4$
Case 2: $a = 2$, $a + 7d = 30 \Rightarrow d = 4$
Sum of first 20 terms:
$$S_{20} = \frac{20}{2}[2a + 19d]$$
- Case 1: $S_{20} = 10[60 + 19(-4)] = 10[60 - 76] = 10 \times (-16) = \mathbf{-160}$
- Case 2: $S_{20} = 10[4 + 19(4)] = 10[4 + 76] = 10 \times 80 = \mathbf{800}$
Source: Arithmetic Progressions, Chapter 5
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Explanation
- The key step is treating $a$ and $a_8$ as two unknowns with known sum (32) and product (60), forming a quadratic equation.
- Both cases are valid — examiners expect both solutions with corresponding $S_{20}$.
- Use $S_n = \frac{n}{2}[2a + (n-1)d]$ correctly for each case.
- Don't forget to verify: $30 \times 2 = 60$ ✓ and $30 + 2 = 32$ ✓.