Question

find the value of Y if the first three terms of an ap respectively are 3 Y - 1, 3 Y + 5 and 5 Y + 1

Answer 1

the answer of the above question is given in the picture

Answer 2

given
a₁=3y-1
a₂=3y+5
a₃=5y+1
now d=a₂-a₁.........(1)
also d=a₃-a₂.........(2)
equating (1) and (2)
a₂-a₁=a₃-a₂
⇒3y5-(3y-1)=5y+1-(3y-5)
⇒3y+5-3y+1=5y+1-3y-5
⇒2y=10
⇒y=5
check
a₁=3(5)-1
a₁=14
a₂=3(5)+5
a₂=20
a₃=5(5)+1
a₃=26
d=a₂-a₁
d=20-14
d₁=6
d=a₃-a₂
d=26-20
d₂=6
thus d₁=d₂
thus y=5 is acceptable
another value y=4 has come out in the above solution that value is not wrong. but if we use y=5
then
a₁=3(4)-1
a₁=11
a₂=3(4)+5
a₂=17
a₃=5(4)+1
a₃=21
d₁=a₂-a₁
d₁=17-11
d₁=6
d₂=21-17
d₂=4
thus d₁≠d₂
thus y=4 is rejected