Which term of the series 1,3,9,27,... is 6561?

Which term of the series 1,3,9,27,... is 6561?

1. 7th
2. 8th
3. 9th
4. 10th

Solution: 3.9th term

This is a geometric progression. 

Let us assume first term=a, common ratio=r, nth term=tₙ

Given, a=1, r=3÷1=3, tₙ=6561

For a geometric progression, tₙ=ar⁽ⁿ⁻¹⁾

(We are to use this formula to find the answer, we know the value of first term(a), common ratio(r) and the nth term. The unknown value is n. We substitute all the values to the formula)

tₙ=ar⁽ⁿ⁻¹⁾

⇒6561=1*3⁽ⁿ⁻¹⁾

⇒6561=3⁽ⁿ⁻¹⁾

 (now we will express 6561 in the power of 3, because in the RHS the value is in power of 3.so by expressing the LHS value in power of 3,we can equate both sides and by the concept of indices we can find the value of n)

⇒3⁸=3⁽ⁿ⁻¹⁾

⇒8=n-1
(if aˣ=aⁿ then x=n)
⇒8+1=n
⇒9=n
⇒n=9.