Question Number 27908 by v7277668420 last updated on 16/Jan/18
![IF a_(1,) a_2 ,.....,a_(n−1) ,a_n are in AP then prove that 1/a_1 .a_n + 1/a_2 .a_(n−1) + 1/a_3 .a_(n−2) +...+1/a_n .a_1 = 2/a_1 +a_(n ) [1/a_(1 ) +1/a_2 +....+1/a_n ]](Q27908.png)
$${IF}\:\:{a}_{\mathrm{1},} {a}_{\mathrm{2}} ,.....,{a}_{{n}−\mathrm{1}} ,{a}_{{n}} \:{are}\:{in}\:{AP}\:{then}\:{prove}\:{that} \\ $$$$\mathrm{1}/{a}_{\mathrm{1}} .{a}_{{n}} +\:\mathrm{1}/{a}_{\mathrm{2}} .{a}_{{n}−\mathrm{1}} +\:\mathrm{1}/{a}_{\mathrm{3}} .{a}_{{n}−\mathrm{2}} +...+\mathrm{1}/{a}_{{n}} .{a}_{\mathrm{1}} = \\ $$$$\mathrm{2}/{a}_{\mathrm{1}} +{a}_{{n}\:} \left[\mathrm{1}/{a}_{\mathrm{1}\:} +\mathrm{1}/{a}_{\mathrm{2}} +....+\mathrm{1}/{a}_{{n}} \right] \\ $$
Commented by v7277668420 last updated on 17/Jan/18
$$ \\ $$$${please}\:{help} \\ $$
Answered by ajfour last updated on 17/Jan/18
![l.h.s.=Σ_(r=1) ^n (1/(a_r a_(n−r+1) )) =Σ_(r=1) ^n (1/([a_1 +(r−1)d][a_1 +(n−r)d])) =(1/(2a_1 +(n−1)d)) Σ_(r=1) ^n (([a_1 +(r−1)d]+[a_1 +(n−r)d])/([a_1 +(r−1)d][a_1 +(n−r)d])) =(1/(2a_1 +(n−1)d)) Σ_(r=1) ^n [(1/(a_1 +(n−r)d))+(1/(a_1 +(r−1)d))] =(2/(a_n +a_1 ))[(1/a_n )+(1/a_(n−1) )+(1/a_(n−2) )+...(1/a_1 )] .](Q27937.png)
$${l}.{h}.{s}.=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{{a}_{{r}} {a}_{{n}−{r}+\mathrm{1}} } \\ $$$$=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\left[{a}_{\mathrm{1}} +\left({r}−\mathrm{1}\right){d}\right]\left[{a}_{\mathrm{1}} +\left({n}−{r}\right){d}\right]} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{a}_{\mathrm{1}} +\left({n}−\mathrm{1}\right){d}}\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left[{a}_{\mathrm{1}} +\left({r}−\mathrm{1}\right){d}\right]+\left[{a}_{\mathrm{1}} +\left({n}−{r}\right){d}\right]}{\left[{a}_{\mathrm{1}} +\left({r}−\mathrm{1}\right){d}\right]\left[{a}_{\mathrm{1}} +\left({n}−{r}\right){d}\right]} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{a}_{\mathrm{1}} +\left({n}−\mathrm{1}\right){d}}\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\frac{\mathrm{1}}{{a}_{\mathrm{1}} +\left({n}−{r}\right){d}}+\frac{\mathrm{1}}{{a}_{\mathrm{1}} +\left({r}−\mathrm{1}\right){d}}\right] \\ $$$$=\frac{\mathrm{2}}{{a}_{{n}} +{a}_{\mathrm{1}} }\left[\frac{\mathrm{1}}{{a}_{{n}} }+\frac{\mathrm{1}}{{a}_{{n}−\mathrm{1}} }+\frac{\mathrm{1}}{{a}_{{n}−\mathrm{2}} }+...\frac{\mathrm{1}}{{a}_{\mathrm{1}} }\right]\:. \\ $$
Commented by v7277668420 last updated on 17/Jan/18
$${any}\:{easy}\:{method} \\ $$