Question

Insert two number between 3 and 81 so that the resulting sequence in gp

Answer 1

a(r raised to the power n-1)=3. here a=3.a(r raised to the power n-1)=81.we know that a=3.so(r raised to the power n-1)=81/3=27.27 can be written as 3 raised to the power 3.that is r=3 and n-1=3.if n-1=3 then n=3+1=4.geometric progression is 3,9,27,81.that a,a(r raised to the power 2-1),a(r raised to the power 3-1),a (r raised to the power 4-1).

Answer 2

$\purple{ \mathtt{ \huge \underline{ \fbox{ \: Solution :</p><p> \: \: \: }}}}$

Let ,

The two numbers are x and y , so

GP is 3 , x , y , 81 and common ratio is r

$\star \: \sf First \: term = a_{1} = 3$

$\sf \star \: Fourth \: term = a {(r)}^{3} = 3 {(r)}^{3} \\ \\ \sf \hookrightarrow 81 = 3 {(r)}^{3} \: \: \: \: \{ \because Fourth \: term = 81\} \\ \\ \sf \hookrightarrow {(r)}^{3} = 27 \\ \\\sf \hookrightarrow r = 3$

Hence , the common ratio is 3

Now ,

$\sf \star \: \: Second \: term = ar = 3(3) = 9 \\ \\ \sf \star \: \: Third \: term = a {(r)}^{2} = 3 {(3)}^{2} = 27</p><p>$

Therefore , The required numbers are 9 and 27