Question

IF (3y-1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y .

Answer 1

AP : $3y-1,3y+5,5y+1.$

As per the question , we have

First term, $a=3y-1$----(1)

Second term , $a_2=3y+5$

$a+d=3y+5$----(2)

Third term,$a_3=5y+1$

$a+2d=5y+1$ -----(3)

Now, subtracting equation (2) from equation(3), we have

$d=2y-4$----(4)

Similarly, subtracting equation (1) from equation(3) , we get

$2d=2y+2$

$d=y+1$( Multiplying the whole equation by $\frac{1}{2}$) -----(5)

Solving equation(4) and (5) , we obtain

$2y-4=y+1$

$2y-y=1+4$

$y=5$

Thus , y = 5