4^th term of a G. P is 8 and 10^th term is 27. Then its 6^th term is?

Concept:

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

• Common ratio = r = \(\frac{{{a_2}}}{{{a_1}}} = \frac{{{a_3}}}{{{a_2}}} = \ldots = \frac{{{a_n}}}{{{a_{n - 1}}}}\)
• n^th  term of the G.P. is a_n = arⁿ⁻¹
• Sum of n terms = s = \(\frac{{a\;\left( {{r^n} - 1} \right)}}{{r - \;1}}\); where r >1
• Sum of n terms = s = \(\frac{{a\;\left( {{r^n} - 1} \right)}}{{r - \;1}}\); where r <1

Calculation:

Given:

4th term of a G. P is 8 and 10th term is 27

nth  term of the G.P. is T_n = a r^n-1

∴ T₄ = a. r³ = 8      ----(1)

T₁₀ = a r⁹ = 27      ----(2)

Equation (2) ÷ (1), we get 

\( {r^6} = \frac{{27}}{8}\)

\(\Rightarrow {\left( {{{\rm{r}}^2}} \right)^3} = {\rm{\;}}{\left( {\frac{3}{2}} \right)^3}\)

∴ \({r^2} = \frac{3}{2}\)

T₆ = a r⁵

= a r³.r²

\(= 8 \times \frac{3}{2} \)