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