To calculate a10, we'll write the formula of the general term of an arithmetical progression:

an = a1 + (n-1)*r

a10 = a1 + (10-1)*r

a10 = 10 + 9r

We don't know yet the ratio r. But we know a4 = 19

a4 = a1 + 3r

19 = 10 + 3r

We'll subtract 10 both sides:

3r = 9

We'll divide by 3:

r = 3

Now, we can calculate a10.

a10 = 10 + 9r

a10 = 10 + 9*3

a10 = 10 + 27

**a10 = 37**

a4 = 19, a1 = 10 to find a10 of the AP.

Solution:

For an Arithmetic progression, the rth term ar = a1+(r-1)d, where d is the common difference. So,

a4 - a1 = a1+3d-a1 = 3d

Here a4 -a1 = 19- 10 = 9

3d =9

d =3.

S0 a10 = a1+(10-1)d = 10 +(10-1)*3 = 37

I believe the initial A.P. used in the question for arithmetic progression, and the terms a1, a4 and a10 refer to the first, fourth and tenth term of the progression.

In an arithmetic progression the difference (d) between any two consecutive terms is same.

Thus the different terms of the progression can be represented as:

a1= a1

a2 = a1 +d

a3 = a1 + 2d

a4 = a1 + 3d

...

an = a1 + (n -1)*d

It is given:

a1 = 10, and a4 = 19

Therefore:

a4 = 19 = a1 + 3d = 10 + 3d

3d = 19 - 10 = 9

Therefore:

d = 9/3 = 3

Therefore:

a10 = a1 + (10 - 1)d

= 10 + 9*3

= 10 + 27 = 37

Answer:

a10 = 37