Question

The first three terms of an AP respectively are 3y – 1, 3y + 5 and 5y + 1. Then y equals​

Answer 1

Step-by-step explanation:

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

3y-1+2(6)

3y-1+12

3y+11

y=-11/3 OK

Answer 2

Answer:-

$\sf{The \ value \ of \ y \ is \ 5.}$

Given:

• The first three terms of an AP are 3y-1, 3y+5 and 5y+1 respectively.

To find:

• The value of y.

Solution:

$\boxed{\sf{2\times \ t_{2}=t_{1}+t_{3}}}$

$\sf{2(3y+5)=(3y-1)+(5y+1)}$

$\sf{\therefore{6y+10=8y}}$

$\sf{\therefore{8y-6y=10}}$

$\sf{\therefore{2y=10}}$

$\sf{\therefore{y=\frac{10}{2}}}$

$\boxed{\sf{\therefore{y=5}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ y \ is \ 5.}}}$