The first three terms pf A.P are $(3 y - 1) . (3 y + 5)$ and $(5 y + 1)$ . Then $Y$ equals to___________.

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Solution

Consider the given value of A.P. $(3 y - 1), (3 y + 5)$ and $5 y + 1$ .

Given that these three term are in A.P. so,

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

$6 = 3 y - 4$

$3 y = 10$

$y = \frac{10}{3}$
Hence this is the the required value of y.

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