Question

if 4,x1,x2,x3,28, are A.p then x3 =?

Answer 1

Answer:

$x_{3} = 16$

Step-by-step explanation:

The given series is-

$4, x_{1}, x_{2}, x_{3}, 28$

First term = 4

Last term = 28

We know that-

The nth term of A.P is-

$T_{n} = a + (n-1)d$

Where a is the first term of A.P and

d is difference between every term.

We have-

$T_{5} = 28$

$4 + (5-1)d = 28$

$4 + 4d = 28$

4d = 28 - 4

4d = 24

$d = \dfrac{24}{4}$

d = 6

So, $T_{4} = a+(4-1)d$

$x_{3} = 4+3 \times 4$

$x_{3} = 4+12$

$x_{3} = 16$