Question

which term of the series 3,√3,1,1/√3....is 1/243?​

Answer 1

Which term of series ; 3, √3 , 1 , 1/√3 ....... is 1/243 ?

solution : here series ; 3 , √3 , 1 , 1/√3 ......

we can see it is geometric progression because ratio of two consecutive terms always remains same. i.e., (√3)/3 = 1/√3 = (1/√3)/1 = 1/√3

now using formula of nth term of geometric progression.

nth term = arⁿ¯¹

here a = 3 and r = 1/√3 and nth term = 243

so, 1/243 = 3(1/√3)ⁿ¯¹

⇒1/729 = (1/√3)ⁿ¯¹

⇒1/729 = (3)^{-(n - 1)/2} [ because 1/√3 = (3)^{-1/2} ]

⇒1/3⁶ = (3)^{(1 - n)/2}

⇒ 3¯⁶ = (3)^{(1 - n)/2}

⇒-6 × 2 = 1 - n

⇒n = 13

Therefore 13th term of the series is 1/243.

Answer 2

Answer:

=1/3^6

Step-by-step explanation:

tn=243

r=t2/t1

√3/3=√3/√3×√3

r=1/√3

3(1/√3)^n-1

=1/3^5

(1/√3)^n-1=1/3^6