Question

The first three terms of an A.P. respectively are 3y - 1, 3y + 5 and 5y + 1. Then, y equals
(a) −3
(b) 4
(c) 5
(d) 2

Answer 1

Answer:

The value of y is 5 .

Among the given options option (c) 5 is a correct answer.

Step-by-step explanation:

Given :  

(3y - 1), (3y + 5) and (5y + 1) are in A.P

Let a1 =  (3y - 1) , a2 = (3y + 5)  , a3 = (5y + 1)

If terms are in A.P, then common difference (d) of any two consecutive terms is same.  

a2 - a1 = a3 - a2

(3y + 5) - (3y - 1) = (5y + 1) - (3y + 5)

3y + 5 - 3y + 1 = 5y + 1 - 3y - 5

6 = 2y - 4

6 + 4 = 2y

10 = 2y

y = 10/2

y = 5

Hence, the value of y is 5 .

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Answer:-

y = 5

Explanation:-

Given

3y-1, 3y+5, 5y+1 are the first three terms of an A.P.

To Find

Value of y

Solution

The series is...

3y-1, 3y+5 , 5y+1 ....

Here

a1 = 3y-1

a2 = 3y+5

a3 = 5y+1

since, the series is in A.P. Common difference(d) is equal!

Common difference

a2-a1 = a3-a2

3y+5 - (3y-1) = 5y+1 - (3y+5)

3y+5-3y+1 = 5y+1-3y-5

6 = 2y-4

2y = 6+4

2y = 10

$\boxed{y = 5}$