Question

The first three terms of an a.p.respectively are 3y-1,3y + 5 and 5y +1. Then.y equal

Answer 1

$Given \: (3y-1), (3y+5) \: and \: (5y+1) \: are$

$first \: three \: terms \: of \: an \: A.P$

$We \: know \: that,$

$\blue { ( If \: a,b\: and \: c \: are \: in \: A.P }$

$\blue { then \: 2b = a + c )}$

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

$\implies 2(3y+5) = 8y$

$\implies 3y+5 = \frac{8y}{2}$

$\implies 3y + 5 = 4y$

$\implies 5 = 4y - 3y$

$\implies 5 = y$

$\implies y = 5$

Therefore.,

$\red{ Value \: of \: y }\green { = 5 }$

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Answer 2

★ Formula used :

• If a, b and c are in A, then 2b = a + c

★ Solution :

2 ( 3y + 5 ) = 3y - 1 + 5y + 1

⇒6y + 10 = 3y + 5y

⇒6y + 10 = 8y

⇒ 10 = 2y

⇒ 10/2 = y

⇒ 5 = y