Question Number 11262 by 786786AM last updated on 18/Mar/17
$$\mathrm{The}\:\mathrm{k}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{of}\:\mathrm{a}\:\mathrm{sequence}\:\mathrm{is}\:\mathrm{K},\:\mathrm{the}\:\mathrm{m}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{of}\:\:\mathrm{M}\:\mathrm{and}\:\mathrm{n}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{is}\:\mathrm{N}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{geometic}, \\ $$$$\left(\mathrm{m}−\mathrm{n}\right)\:\mathrm{log}\:\mathrm{K}\:+\:\left(\mathrm{n}−\mathrm{k}\right)\:\mathrm{log}\:\mathrm{M}\:+\:\left(\mathrm{k}−\mathrm{m}\right)\:\mathrm{log}\:\mathrm{N}\:=\:\mathrm{0}.\: \\ $$
Answered by mrW1 last updated on 20/Mar/17
![K=a∙q^(k−1) M=a∙q^(m−1) N=a∙q^(n−1) (m−n)log K=(m−n)[log a+(k−1)log q] (n−k)log M=(n−k)[log a+(m−1)log q] (k−m)log K=(k−m)[log a+(n−1)log q] S=(m−n+n−k+k−m)log a+[(m−n)(k−1)+(n−k)(m−1)+(k−m)(n−1)]log q =[km−kn−m+n+mn−km−n+k+kn−mn−k+m]log q =0](Q11331.png)
$${K}={a}\centerdot{q}^{{k}−\mathrm{1}} \\ $$$${M}={a}\centerdot{q}^{{m}−\mathrm{1}} \\ $$$${N}={a}\centerdot{q}^{{n}−\mathrm{1}} \\ $$$$\left({m}−{n}\right)\mathrm{log}\:{K}=\left({m}−{n}\right)\left[\mathrm{log}\:{a}+\left({k}−\mathrm{1}\right)\mathrm{log}\:{q}\right] \\ $$$$\left({n}−{k}\right)\mathrm{log}\:{M}=\left({n}−{k}\right)\left[\mathrm{log}\:{a}+\left({m}−\mathrm{1}\right)\mathrm{log}\:{q}\right] \\ $$$$\left({k}−{m}\right)\mathrm{log}\:{K}=\left({k}−{m}\right)\left[\mathrm{log}\:{a}+\left({n}−\mathrm{1}\right)\mathrm{log}\:{q}\right] \\ $$$${S}=\left({m}−{n}+{n}−{k}+{k}−{m}\right)\mathrm{log}\:{a}+\left[\left({m}−{n}\right)\left({k}−\mathrm{1}\right)+\left({n}−{k}\right)\left({m}−\mathrm{1}\right)+\left({k}−{m}\right)\left({n}−\mathrm{1}\right)\right]\mathrm{log}\:{q} \\ $$$$=\left[{km}−{kn}−{m}+{n}+{mn}−{km}−{n}+{k}+{kn}−{mn}−{k}+{m}\right]\mathrm{log}\:{q} \\ $$$$=\mathrm{0} \\ $$