Question

The 3 arithmetic means between 3 and 19 are:

• A. 7, 11, 15
• B. 5, 10, 15
• C. 5, 7, 9
• D. 6, 11, 16

Hint:

Inserting $n$ arithmetic means between two numbers creates an AP with $(n+2)$ total terms. Use the common difference formula $d = \frac\textlast - \textfirst\texttotal terms - 1$ to solve quickly.

Answer 1

The Correct Option is A

We are given two numbers: $a_1 = 3$ and $a_5 = 19$.
We need to insert 3 arithmetic means between them, which means the sequence will have 5 terms in total:
\[ 3, \, A_1, \, A_2, \, A_3, \, 19 \] Since it is an arithmetic progression (AP), the common difference $d$ is given by:
\[ d = \frac\textlast term - \textfirst term\textnumber of terms - 1 \] Substituting:
\[ d = \frac19 - 35 - 1 = \frac164 = 4 \] Thus, the AP is:
First term = 3
Second term = $3 + 4 = 7$
Third term = $7 + 4 = 11$
Fourth term = $11 + 4 = 15$
Fifth term = $15 + 4 = 19$
Therefore, the 3 arithmetic means are $\mathbf7, 11, 15$.