If we have to insert 4 arithmetic means between 10 and 15, we can take the nth term of the AP to be a+(r-1)d.

Here a=10 is the first term.

And as 15 is the 6th term, a+5d=15, or 10+5d=15 or 5d=15-10=5.

d=1

Therefore the 4 terms between 10 and 15 are 10+1, 10+2*1,10+3*1 and 10+4*1.

The four required numbers are 11, 12, 13 and 14.

To in sert 4 a.m terms between 10 and 15.

We assume that 10 and 15 are 1st and 6th term of an arith matic series , whose nth term ,an = a1+(n-1)d, where a1 = 1st yerm, d is the common difference.

So a1 = 10

a6 = a1+(6-1)d =15

Therefore a6-a1 = (a1+5d) -a1 = 5d = 15-10= 5, d = 5/5 =1 is the common difference.

Therefore the 4 consecutive terms after a1 are 10+1*1, 10+2*1, 10+3*1and 10+4*1. Or 11 ,12, 13 and 14.

We'll note a2,a3,a4 and a5 the arithmetic means inserted between the terms 10 and 15. The sequence resulted is:

10, a2, a3, a4,a5, 15,...

This sequence is an arithemtic progression, so the common difference is:

a2-10 = a3-a2 = a4-a3 = a5-a4 = 15-a5

The first term is a1 = 10 and a6 = 15.

a6 = a1 + 5d

15 = 10 + 5d

5d = 15-10

5d = 5

d = 1

a2 = a1+d

a2 = 10 +1

a2 = 11

a3 = a1 + 2d

a3 = 10 + 2

a3 = 12

a4 = a1 + 3d

a4 = 10 + 3

a4 = 13

a5 = a1 + 4d

a5 = 10 + 4

a5 = 14

**The arithmetic means inserted between 10 and 15 are :**

**a2 = 11**

**a3 = 12**

**a4 = 13**

**a5 = 14**