Question

2,x,26 are in AP.find the value of x

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore x=14}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline \bold{Given : } \\ \implies 2,x,26 \: are \: in \:A.P \\ \\ \underline \bold{To \: Find : } \\ \implies x = ?$

• According to given question :

$\bold{We \: know \: that \to} \\ \bold{If \: a,b,c \: are \: in \: A.P \: So,} \\ \bold{\boxed{2b = a + c}} \\ \\ \bold{a = 2} \\ \bold{b = x} \\ \bold{c =26} \\ \\ \bold{Putting \: given \: values : } \\ \implies 2 \times x = 2 + 26 \\ \\ \implies 2x = 28 \\ \\ \implies x = \frac{ \cancel{28}}{ \cancel2} \\ \\ \bold{\implies x = 14}$

Answer 2

Answer:

$\bold\red{x=14}$

Step-by-step explanation:

It is being given that,

2, x, 26 are in AP.

Now, we know that,

first term, a = 2

Let, the common difference be 'd'

So, we have,

2nd term = 1st term + common difference

=> x = 2 + d .........(i)

Also,

3rd term = 2nd term + common difference

=> 26 = x + d

Putting the value of x from eqn (i),

we get,

=> 26 = (2+d) +d

=> 26 = 2+ 2d

=> 1+ d = 13

=> d = 13 -1

=> d = 12

Therefore,

2nd term , x = 2 + d = 2+ 12 = 14

Hence, x = 14

Alternative Method :

When 3 terms are given in AP, then,

$2nd\:term = \frac{1st\:term+3rd\:term}{2}$

So, here

$=>2nd\:term = x= \frac{2+26}{2}$

$=>2nd\:term = x= \frac{28}{2}$

Hence, x = 14