Mathematics — Standard (041) — CBSE Class 10 board question

Step 1: Use the condition that 15th term exceeds 8th term by 21.

$a_{15} - a_8 = 21$

$[a + 14d] - [a + 7d] = 21$

$7d = 21 \Rightarrow d = 3$

Step 2: Use $S_{10} = 55$.

$S_{10} = \dfrac{10}{2}[2a + 9d] = 55$

$5[2a + 27] = 55$

$2a + 27 = 11$

$2a = -16 \Rightarrow a = -8$

Step 3: Form the A.P.

$a = -8,\ d = 3$

A.P.: –8, –5, –2, 1, 4, 7, ...

Source: Chapter 5 (Arithmetic Progressions), Exercise 5.2 & 5.3

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• Examiners award 1 mark for finding $d$, 1 mark for finding $a$, and 1 mark for writing the A.P.
• Use $a_n = a + (n-1)d$ and $S_n = \frac{n}{2}[2a + (n-1)d]$ — both standard formulas from the chapter.
• Always write at least 4–5 terms of the A.P. at the end to confirm the answer.