Question

The 3rd term of g.P is 2/3 and the 6th term is 2/18 , then the 1st term is

Answer 1

First term of the GP is $\sqrt[3]{36}(\frac{2}{3} )$.

• 3rd term of GP is given as 2/3.
• 6th term of GP is 2/18.
• nth term of GP is $ar^{n-1}$.

$t_3 = ar^{2}$ = 2/3 .....(Equation 1)

$t_6 = ar^{5}$ = 2/18 .....(Equation 2)

• Dividing equation 2 by 1 , we get

$r^{3}  = \frac{1}{6}$

r = $(\frac{1}{6} )^{\frac{1}{3} }$

• Now substituting the value of r in equation 1 we get

$ar^{2}$ = 2/3

$a{(\frac{1}{6} )^{\frac{1}{3} }}^{2}$ = 2/3

a = $\sqrt[3]{36}(\frac{2}{3} )$

• First term of the GP is $\sqrt[3]{36}(\frac{2}{3} )$.