Question

the first three terms of an AP are 3y-1, 3y+5, 5y+1 respectively​

Answer 1

$\Large\bf\red{Given :- }$

The terms (3y – 1), (3y + 5) and (5y + 1) are in AP.

$\Large\bf\blue{To\:Find:-}$

The value of y

$\Large\bf\green{Solution:-}$

The given 3 terms are in A.P

So,$\bf a_2-a_1=a_3-a_2$

$\bf \implies (3y+5)-(3y-1)=(5y+5)-(3y+5)$

$\bf \implies 3y+5-3y-1=5y+5-3y+5$

$\bf \implies 6=2y-4$

$\bf \implies 2y=10$

$\bf \implies y={\cancel\dfrac{10}{2}}$

${\bf{\underline {\boxed{\pink{y=5}}}}}$

Hence,the value of y is 5.

Answer 2

above answer is correct ✔✔✔