Mathematics — CBSE Class 10 board question

Given: $S_n = 2n^2 + 13n$

Finding the nth term:

$$a_n = S_n - S_{n-1}$$

$$S_{n-1} = 2(n-1)^2 + 13(n-1) = 2n^2 - 4n + 2 + 13n - 13 = 2n^2 + 9n - 11$$

$$a_n = (2n^2 + 13n) - (2n^2 + 9n - 11)$$

$$\boxed{a_n = 4n + 11}$$

Finding the 10th term:

$$a_{10} = 4(10) + 11 = 40 + 11 = \boxed{51}$$

Source: Chapter 5, Exercise 5.3

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• The key formula is $a_n = S_n - S_{n-1}$, valid for $n \geq 2$. Examiners expect you to show this substitution step clearly.
• Always simplify $S_{n-1}$ carefully by expanding $(n-1)^2$ and $(n-1)$.
• Finally substitute $n = 10$ in $a_n$ to get the 10th term. Write both results clearly boxed or underlined for the examiner.