Question Number 18022 by ibraheem160 last updated on 13/Jul/17
![The first term of an A.P is log^a and second term is log^b .show that the sum of first n terms in (1/2)log[(b^(n(n−1)) /a^(n(−3)) )]](Q18022.png)
$${The}\:{first}\:{term}\:{of}\:{an}\:{A}.{P}\:\:{is}\:{log}^{{a}} \:{and}\:{second}\:{term}\:{is}\: \\ $$$${log}^{{b}} .{show}\:{that}\:{the}\:{sum}\:{of}\:{first}\:{n}\:{terms}\:{in}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{log}\left[\frac{{b}^{{n}\left({n}−\mathrm{1}\right)} }{{a}^{{n}\left(−\mathrm{3}\right)} }\right] \\ $$
Answered by prakash jain last updated on 14/Jul/17
![first term=a_1 =log a d=log b−log a S_n =(n/2)(2a_1 +(n−1)d) =(n/2)(2log a+(n−1)(log b−log a)) =(1/2)(2nlog a+n(n−1)log b−n(n−1)log a) =(1/2)(log a^(2n) +log b^(n(n−1)) −log a^(n(n−1)) ) =(1/2)(log b^(n(n−1)) −[log a^(n(n−1)) −log a^(2n) ] =(1/2)(log b^(n(n−1)) −[log(a^(n^2 −n) /a^(2n) )]) =(1/2)(log b^(n(n−1)) −log a^(n(n−3)) ) =(1/2)log (b^(n(n−)) /a^(n(n−3)) )](Q18068.png)
$${first}\:{term}={a}_{\mathrm{1}} =\mathrm{log}\:{a} \\ $$$${d}=\mathrm{log}\:{b}−\mathrm{log}\:{a} \\ $$$${S}_{{n}} =\frac{{n}}{\mathrm{2}}\left(\mathrm{2}{a}_{\mathrm{1}} +\left({n}−\mathrm{1}\right){d}\right) \\ $$$$=\frac{{n}}{\mathrm{2}}\left(\mathrm{2log}\:{a}+\left({n}−\mathrm{1}\right)\left(\mathrm{log}\:{b}−\mathrm{log}\:{a}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{n}\mathrm{log}\:{a}+{n}\left({n}−\mathrm{1}\right)\mathrm{log}\:{b}−{n}\left({n}−\mathrm{1}\right)\mathrm{log}\:{a}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{log}\:{a}^{\mathrm{2}{n}} +\mathrm{log}\:{b}^{{n}\left({n}−\mathrm{1}\right)} −\mathrm{log}\:{a}^{{n}\left({n}−\mathrm{1}\right)} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{log}\:{b}^{{n}\left({n}−\mathrm{1}\right)} −\left[\mathrm{log}\:{a}^{{n}\left({n}−\mathrm{1}\right)} −\mathrm{log}\:{a}^{\mathrm{2}{n}} \right]\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{log}\:{b}^{{n}\left({n}−\mathrm{1}\right)} −\left[\mathrm{log}\frac{{a}^{{n}^{\mathrm{2}} −{n}} }{{a}^{\mathrm{2}{n}} }\right]\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{log}\:{b}^{{n}\left({n}−\mathrm{1}\right)} −\mathrm{log}\:{a}^{{n}\left({n}−\mathrm{3}\right)} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\frac{{b}^{{n}\left({n}−\right)} }{{a}^{{n}\left({n}−\mathrm{3}\right)} } \\ $$