Insert 5 arithmetic means between 3 and 27

Question

Insert 5 arithmetic means between 3 and 27.

Solution — Step by Step

We need to insert 5 numbers between 3 and 27 such that all 7 numbers (3, the 5 means, and 27) form an arithmetic progression (AP).

So the sequence looks like: $3, a_1, a_2, a_3, a_4, a_5, 27$

This is an AP with 7 terms: first term $a = 3$ , last term $l = 27$ , and total number of terms $n = 7$ .

For an AP: last term $= a + (n-1)d$

27 = 3 + (7-1) \cdot d

27 = 3 + 6d

6d = 24

d = 4

Starting from 3 and adding $d = 4$ each time:

a_1 = 3 + 4 = 7

a_2 = 7 + 4 = 11

a_3 = 11 + 4 = 15

a_4 = 15 + 4 = 19

a_5 = 19 + 4 = 23

The full AP is: $3, 7, 11, 15, 19, 23, 27$

Check: Last term $= 3 + 6 \times 4 = 3 + 24 = 27$ ✓

The 5 arithmetic means are: 7, 11, 15, 19, 23.

Why This Works

When we insert $m$ arithmetic means between two numbers $a$ and $b$ , we’re really just building an AP with $m + 2$ total terms (the two given numbers plus the $m$ means). The common difference $d$ is determined by:

d = \frac{b - a}{m + 1}

Here: $d = \frac{27 - 3}{5 + 1} = \frac{24}{6} = 4$ .

This formula is the quickest path in an exam — memorise it.

d = \frac{b - a}{m + 1}

The $k$ -th arithmetic mean $= a + kd$ , for $k = 1, 2, \ldots, m$ .

Alternative Method

You can also notice that the 5 arithmetic means form the 2nd through 6th terms of the AP. Using the general term formula $T_n = a + (n-1)d$ with $a = 3$ and $d = 4$ :

T_2 = 3 + 4 = 7, \quad T_3 = 11, \quad T_4 = 15, \quad T_5 = 19, \quad T_6 = 23

Common Mistake

The most common error is treating the total number of terms as 5 (the number of means) rather than 7 (the two given values + 5 means). This leads to $d = \frac{27-3}{4} = 6$ , giving the wrong sequence 3, 9, 15, 21, 27 — which has only 3 arithmetic means, not 5. Always count total terms = number of means + 2.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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