1.

Find: the 12th term of the G.P. \(\frac{1}{{a^3 x^3}}\) = a4x4

Answer»

Tn = arn-1

a = \(\frac{1}{a^3x^3}\), r = \(\cfrac{ax}{\frac{1}{a^3x^3}}\) = \({a^4}{x^4}\)

 \(\therefore\) T12 = \(\frac{1}{a^3x^3}(a^4 x^4)^{12-1}\)

\(\frac{1}{a^3x^3}(a^4x^4)^{11}\)

= a41x41

∴ The 12 term is a41 x41.



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