Question

If 1/3, y1, y2, 9 are in G.P. then the value of y2 is ………

Answer 1

$Given \:G.P : \frac{1}{3} , y_{1} , y_{2}, 9$

$First\:term = a = a_{1} = \frac{1}{3}$

$Let \: common \:ratio = r$

$a_{4} = 9$

$\implies ar^{3} = 9$

$\boxed{\pink{ \because n^{th} \:term (a_{n}) = ar^{n-1} }}$

$\implies \frac{1}{3} r^{3} = 9$

$\implies r^{3} = 9 \times 3$

$\implies r^{3} = 27$

$\implies r^{3} = 3^{3}$

$\implies r = 3$

$Now, \blue{ y_{1}} = a_{2}$

$= ar$

$= \frac{1}{3} \times 3$

$= 1$

$\blue{ y_{2}} = a_{3}$

$= ar^{2}$

$= \frac{1}{3} \times 3^{2}$

$= 3$

Therefore.,

$\red{ Value \: of \: y_{1} } \green{=1}$

$\red{ Value \: of \: y_{2} } \green{=3}$

•••♪