Which term of the following sequences:(a) `2,2sqrt(2),4,`. . . is 128? (b) `sqrt(3),3,3sqrt(3),`. . . is 729?(c) `1/3,1/9,1/(27),`. . . is `1/(19683)?`

Answer

N/a

Explanation

(a) Given sequence : `2,2sqrt(2),4….`
First term `a_(1)=a=2` and 2nd term `a_(2)=2sqrt(2),` then
`"Common ratio "r=(a_(2))/(a)=(2sqrt(2))/(2)=sqrt(2)`
`"Let "a_(n)=128`
`therefore" "ar^(n-1)=128 rArr 2.(sqrt(2))^(n-1)=128`
`rArr" "(2)^((n-1)/(2))=64`
`therefore" "(2)^((n-1)/(2))=(2)^(6)rArr(n-1)/(2)=6,`
`rArr" "n=13`
Therefore, 128 is the 13th term of the sequence.
(b) Given sequence : `sqrt(3),3,3sqrt(3)....`
First `a=sqrt(3)` and 2nd term `a_(2)=3,` then
`"Common ratio "r=(a_(2))/(a)=(3)/(sqrt(3))=sqrt(3)`
`"Let "a_(n)=729`
`therefore" "ar^(n-1)=729`
`therefore" "sqrt(3)cdot(sqrt(3))^(n-1)=729 rArr (sqrt(3))^(n)=729`
`rArr" "(3)^((n)/(2))=3^(6) rArr" "(n)/(2)=6`
`rArr" "n=12`
Therefore, 729 is the 12th term of the sequence.
(c) Given sequence : `(1)/(3),(1)/(9),(1)/(27),....`
First term `a=(1)/(3)` and 2nd term `a_(2)=(1)/(9),` then
`"Common ratio "r=(a_(2))/(a)=(1//9)/(1//3)=(1)/(3)`
`"Let "a_(n)=(1)/(19683)`
`therefore" "ar^(n-1)=(1)/(19683)`
`therefore" "((1)/(3))((1)/(3))^(n-1)=(1)/(19683)`
`rArr" "((1)/(3))^(n)=((1)/(3))^(9)rArrn=9`
Therefore, `(1)/(19683)` is the 9th term of the progression.