# Which term of the GP 3, 9, 27, 81, ____ is 6561?

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

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Correct Answer - Option 3 : 8th

Concept:

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

• Common ratio = r = $\frac{{{{\rm{a}}_2}}}{{{{\rm{a}}_1}}} = \frac{{{{\rm{a}}_3}}}{{{{\rm{a}}_2}}} = \ldots = \frac{{{{\rm{a}}_{\rm{n}}}}}{{{{\rm{a}}_{{\rm{n}} - 1}}}}$
• nth  term of the G.P. is an = 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 8th term of the given sequence.

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