Question

The first three terms of an AP respectively are 2y-1, 2y + 3 and 4y +3. Then value of y

is……………​

Answer 1

Given,

The first three terms of an AP respectively are 2y-1, 2y + 3, and 4y +3.

To Find,

The value of y =?

Solution,

We can solve the question as follows:

It is given that the first three terms of an AP respectively are 2y-1, 2y + 3, and 4y +3. We have to find the value of y.

In an A.P, the difference between any two consecutive terms is always a constant and is called the common difference.

Therefore, in the given A.P.,

$(2y + 3) - (2y - 1) = (4y + 3) - (2y + 3)$

Now, we will solve the above equation to find the value of y.

$2y + 3- 2y + 1 = 4y + 3 - 2y - 3$

$4y + 4 = 2y$

Taking the y terms to one side,

$4 = 2y - 4y$

$4 = -2y$

$y = -\frac{4}{2} = -2$

Hence, the value of y is equal to -2.

Answer 2

The value of y = 2

Given :

The first three terms of an AP respectively are 2y - 1 , 2y + 3 and 4y + 3

To find :

The value of y

Solution :

Step 1 of 2 :

Form the equation to find the value of y

Here it is given that first three terms of an AP respectively are 2y - 1 , 2y + 3 and 4y + 3

We know if a , b , c are three consecutive terms of an AP then 2b = a + c

Thus we get

$\displaystyle \sf{ 2(2y + 3) = (2y - 1) + (4y + 3) }$

Step 2 of 2 :

Find the value of y

$\displaystyle \sf{ 2(2y + 3) = (2y - 1) + (4y + 3) }$

$\displaystyle \sf{ \implies 4y + 6 = 2y - 1 + 4y + 3}$

$\displaystyle \sf{ \implies 4y + 6 = 6y + 2}$

$\displaystyle \sf{ \implies 4y - 6y = 2 - 6}$

$\displaystyle \sf{ \implies - 2y = - 4}$

$\displaystyle \sf{ \implies y = \frac{ - 4}{ - 2} }$

$\displaystyle \sf{ \implies y = 2 }$

Hence the required value of y = 2

Observation : Taking y = 2 the values obtained are 3 , 7 , 11 which clearly are in AP with common difference as 4

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