Question

If the common difference of an AP is 3 then the difference between 20th terms and 15th terms (\(a_{20} - a_{15}\)) is :

• A. 5
• B. 3
• C. 15
• D. 20

Hint:

For an AP, the difference between the \(m^{th}\) term and the \(n^{th}\) term is given by: \(a_m - a_n = (m-n)d\), where \(d\) is the common difference. Here, we want \(a_{20} - a_{15}\). So, \(m=20\) and \(n=15\). Difference = \((20-15)d = 5d\). Given common difference \(d=3\). Difference = \(5 \times 3 = 15\).

Answer 1

The Correct Option is C

Concept: In an Arithmetic Progression (AP), the \(n^{th}\) term (\(a_n\)) is given by the formula \(a_n = a_1 + (n-1)d\), where \(a_1\) is the first term and \(d\) is the common difference. Step 1: Write expressions for the 20th term (\(a_{20}\)) and the 15th term (\(a_{15}\)) Let \(a_1\) be the first term of the AP. Given: Common difference, \(d = 3\). For the 20th term (\(n=20\)): \[ a_{20} = a_1 + (20-1)d = a_1 + 19d \] Substitute \(d=3\): \[ a_{20} = a_1 + 19(3) = a_1 + 57 \] For the 15th term (\(n=15\)): \[ a_{15} = a_1 + (15-1)d = a_1 + 14d \] Substitute \(d=3\): \[ a_{15} = a_1 + 14(3) = a_1 + 42 \] Step 2: Calculate the difference \(a_{20} - a_{15}\) \[ a_{20} - a_{15} = (a_1 + 19d) - (a_1 + 14d) \] \[ a_{20} - a_{15} = a_1 + 19d - a_1 - 14d \] The \(a_1\) terms cancel out: \[ a_{20} - a_{15} = 19d - 14d \] \[ a_{20} - a_{15} = (19-14)d \] \[ a_{20} - a_{15} = 5d \] Step 3: Substitute the value of the common difference (\(d\)) Given \(d=3\): \[ a_{20} - a_{15} = 5 \times 3 \] \[ a_{20} - a_{15} = 15 \] The difference between the 20th and 15th terms is 15. This matches option (3).