Question

Insert 4 Gm.s between 2 and 486

Answer 1

2, _, _, _, _, _ 486

Answer 2

the question is simply asking us to construct a geometric sequence with :

first term = 2,
sixh term = 486

Answer 3

Yes , is the common ration of any GP =
\[\huge [\frac{ b }{ a }]^{\frac{ 1 }{ n+1 }}\]

Answer 4

i don't like memorizing a formula for such a simple thing, but the formula would be :

\[\huge [\frac{ b }{ a }]^{\frac{ 1 }{ n\color{Red}{-}1 }}\]

Answer 5

in book it is given n+1

Answer 6

lets derive it :)

Answer 7

Okay...

Answer 8

we're given below :
first term = a
nth term = b

Answer 9

nth term of geometric sequence is given by :
\[\large b = ar^{n-1}\]

Answer 10

oh got it

Answer 11

isolate \(\large r\) ^^

Answer 12

once you have the value of \(r\), you can find the 2nd,3rd,4th and 5th terms

Answer 13

so it is just manipulating terms , but the answer is
6 ,18 , 54 ,162

Answer 14

thats right !

Answer 15

first term = 2
6th term = 486

so,

\[\large 486 =2*r^{6-1}\]

Answer 16

\[\large 243 =r^{5}\]

Answer 17

\[\large r=3\]

Answer 18

multiply the first term by \(3\) to get second term
multiply the second term by \(3\) to get third term

and so on...