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

Given AP: $-\dfrac{11}{2}, -3, -\dfrac{1}{2}, \ldots$

Here, $a = -\dfrac{11}{2}$, $d = -3 - \left(-\dfrac{11}{2}\right) = -3 + \dfrac{11}{2} = \dfrac{5}{2}$

Let $a_n = \dfrac{49}{2}$

Using $a_n = a + (n-1)d$:

$$\frac{49}{2} = -\frac{11}{2} + (n-1)\cdot\frac{5}{2}$$

$$\frac{49}{2} + \frac{11}{2} = (n-1)\cdot\frac{5}{2}$$

$$30 = (n-1)\cdot\frac{5}{2} \implies n-1 = 12 \implies n = 13$$

$\therefore \dfrac{49}{2}$ is the 13th term of the AP.

Source: Chapter 5, Exercise 5.2

---

• Identify $a$ and $d$ first; here both involve fractions, so be careful with subtraction.
• Set $a_n$ equal to the given value and solve for $n$.
• Examiners award 1 mark for correct $a$ and $d$, and 1 mark for the correct value of $n$. Show all steps clearly.