Question

Which term of the GP 3, 9, 27, 81, ____ is 6561?
1. 6^th
2. 7^th
3. 8^th
4. 9^th
5. None of these

Answer 1

Correct Answer - Option 3 : 8^th

Concept:

Let us consider sequence a₁, a₂, a₃ ..... a_n is a G.P.

• Common ratio = r = \(\frac{{{{\rm{a}}_2}}}{{{{\rm{a}}_1}}} = \frac{{{{\rm{a}}_3}}}{{{{\rm{a}}_2}}} = \ldots = \frac{{{{\rm{a}}_{\rm{n}}}}}{{{{\rm{a}}_{{\rm{n}} - 1}}}}\)
• n^th  term of the G.P. is a_n = arⁿ⁻¹

 

Calculation:

Given: The sequence 3, 9, 27, 81, ..... is a GP.

Here, we have to find which term of the given sequence is 6561.

As we know that, the the general term of a GP is given by: \(\rm {a_n} = a{r^{n\; - \;1}}\)

Here, a = 3, r = 3 and let an = 6561

⇒ 6561 = (3) ⋅ (3)n - 1

⇒ 3n = 3⁸ 

∴ n = 8

Hence, 6561 is the 8^th term of the given sequence.

• Sum of n terms of GP = s_n = \(\frac{{{\rm{a\;}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - {\rm{\;}}1}}\); where r >1
• Sum of n terms of GP = s_n = \(\frac{{{\rm{a\;}}\left( {1 - {\rm{\;}}{{\rm{r}}^{\rm{n}}}} \right)}}{{1 - {\rm{\;r}}}}\); where r <1
• Sum of infinite GP = \({{\rm{s}}_\infty } = {\rm{\;}}\frac{{\rm{a}}}{{1{\rm{\;}} - {\rm{\;r}}}}{\rm{\;}}\) ; |r| < 1