Mathematics — CBSE Class 10 board question

Given: AP has 40 terms, $S_9 = 153$, sum of last 6 terms = 687.

Step 1: Using $S_9 = 153$

$$S_9 = \frac{9}{2}[2a + 8d] = 153$$
$$2a + 8d = 34 \implies a + 4d = 17 \quad \text{...(i)}$$

Step 2: Sum of last 6 terms = 687

Sum of last 6 terms = $S_{40} - S_{34} = 687$

$$S_{40} = \frac{40}{2}[2a + 39d] = 20(2a + 39d)$$
$$S_{34} = \frac{34}{2}[2a + 33d] = 17(2a + 33d)$$

$$20(2a + 39d) - 17(2a + 33d) = 687$$
$$40a + 780d - 34a - 561d = 687$$
$$6a + 219d = 687 \implies 2a + 73d = 229 \quad \text{...(ii)}$$

Step 3: Solving (i) and (ii)

From (i): $a = 17 - 4d$. Substituting in (ii):
$$2(17 - 4d) + 73d = 229$$
$$34 + 65d = 229$$
$$d = 3$$
$$a = 17 - 12 = 5$$

First term $a = 5$, common difference $d = 3$

Step 4: Sum of all 40 terms

$$S_{40} = \frac{40}{2}[2(5) + 39(3)] = 20[10 + 117] = 20 \times 127 = \mathbf{2540}$$

Source: Chapter 5, Section 5.3 (Sum of first n terms of an AP)

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• Key formula used: $S_n = \frac{n}{2}[2a + (n-1)d]$
• Examiner tip: "Sum of last 6 terms" must be expressed as $S_{40} - S_{34}$ — this is the most common error students make.
• Solve the two linear equations carefully; substitution is the safest method here.
• All three results (a, d, and $S_{40}$) must be stated clearly to score full marks. Each step carries marks, so show all working.