Question

If $a_n$ represents $n^{\text{th}}$ term of the A.P. $-\frac{15}{4}, -\frac{10}{4}, -\frac{5}{4}, \dots$ then value of $a_{16} - a_{12}$ is

• A. $4$
• B. $\frac{5}{4}$
• C. $5$
• D. $\frac{25}{4}$

Hint:

For any A.P., the difference between any two terms $a_p$ and $a_q$ is given by $(p - q)d$.
Here, $a_{16} - a_{12} = (16 - 12)d = 4d$. You do not need to calculate the actual values of $a_{16}$ and $a_{12}$!

Answer 1

The Correct Option is C

Step 1: Find the first term and common difference.
$a_1 = -\frac{15}{4}$, and $d = -\frac{10}{4} - \left(-\frac{15}{4}\right) = \frac{5}{4}$.
Step 2: Compute the two required terms directly, instead of using a shortcut formula. \[ a_{12} = a_1 + 11d = -\frac{15}{4} + \frac{55}{4} = \frac{40}{4} = 10 \] \[ a_{16} = a_1 + 15d = -\frac{15}{4} + \frac{75}{4} = \frac{60}{4} = 15 \]
Step 3: Subtract the two actual term values. \[ a_{16} - a_{12} = 15 - 10 = 5 \]
\[ \boxed{5} \]