Question

if (Y-1),(Y+1)and (2Y+3) are in a.p then value of 'Y'is
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Answer 1

Solution,

Given,

In A.P. = (y-1), (y+1),(2y+3)

Y=?

Then,

According  to the given question,

$t_1= y-1\\t_2=y+1\\t_3=2y+3$

now,

:. Common difference(d)= $t_2-t_1$

                                  = y+1-(y-1)

                                = y+1-y+1

                               = 2

Then, using formula to find $n^{th}$ term of A.P., we get;

$t_n= t1+(n-1)d\\or, t_3= y-1+(3-1).2\\or, 2y+3=y-1+4\\or, 2y-y=-1+4-3\\:. y=0$

Now,

:. A.P.= y-1,y+1,2y+3

= 0-1,0+1,2.0+3

= -1,1,3