Mathematics — CBSE Class 10 board question

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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• 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 $30 + 2 = 32$ ✓.