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:

Avoid calculating individual terms like \(a_{16}\) and \(a_{12}\) separately. Directly use the property \(a_n - a_m = (n-m)d\) to save time in MCQs.

Answer 1

The Correct Option is C

The given sequence is an arithmetic progression (A.P.) with the first term \(a_1 = -\frac{15}{4}\) and the second term \(a_2 = -\frac{10}{4}\). In an A.P., the difference between consecutive terms, known as the common difference \(d\), is constant.

To find the common difference \(d\), we calculate:

\(d = a_2 - a_1 = -\frac{10}{4} - \left(-\frac{15}{4}\right) = \frac{5}{4}\)

Now, we use the formula for the \(n\)-th term of an A.P., which is:

\(a_n = a_1 + (n - 1) \cdot d\)

To find \(a_{16}\), substitute \(n = 16\):

\(a_{16} = -\frac{15}{4} + (16 - 1) \cdot \frac{5}{4}\)

\(a_{16} = -\frac{15}{4} + 15 \cdot \frac{5}{4}\)

\(a_{16} = -\frac{15}{4} + \frac{75}{4} = \frac{60}{4} = 15\)

Next, to find \(a_{12}\), substitute \(n = 12\):

\(a_{12} = -\frac{15}{4} + (12 - 1) \cdot \frac{5}{4}\)

\(a_{12} = -\frac{15}{4} + 11 \cdot \frac{5}{4}\)

\(a_{12} = -\frac{15}{4} + \frac{55}{4} = \frac{40}{4} = 10\)

Now, we compute \(a_{16} - a_{12}\):

\(a_{16} - a_{12} = 15 - 10 = 5\)

Thus, the value of \(a_{16} - a_{12}\) is 5.