Question

if (a+1), 3a and (4a+2) be any three
Consecutive terms of an AP, fine the
value of a.​

Answer 1

Step-by-step explanation:

given

a+1 , 3a and 4a + 1 is in A.P.

So what's the 5th term ?

♦ Now we should first know what is A.P.

» A.P. stands for Arithmetic Progression .

It is type of series in which terms of it increases or decreases with a constant value.

Now as we have

Where

a = first term

d = common difference between two teams.

Now in our A.P

Common difference

= 3a - (a + 1)

= 2a - 1

So we have got d

first we need to check out with the third term

3rd term => 3a + (3-1)(2a - 1) = 4a + 2

=> a+1 + (2)(2a - 1) = 4a + 2

=> a+1 + 4a - 2 = 4a + 2

=> 5a - 1 = 4a + 2

=> a = 3

Now our fifth term

= a+1 + (5-1)(2a-1)

= a + 1 + (4)(2a - 1)

= a + 1 + 8a - 4

= 9a - 3

As a = 3 , So our fifth term is

= 8 a

hope it's help u ✌✌

Answer 2

ANSWER:

• Value of a is 3.

GIVEN:

• (a+1) , 3a , (4a+2) are in A.P

TO FIND:

• Value of 'x'

SOLUTION:

We know that common difference in A.P is same.

= (a+1) , (3a) , (4a+2)

Now :

=> 3a-a-1 = 4a+2-3a

=> 2a-1 = a+2

=> 2a-a = 2+1

=> a = 3

Value of a is 3.

NOTE:

• We know that in Arithmetic progression Common difference is same. Using this property we can find the value of a.
• Simplify the equation and find the value of the variable.