Question Number 58248 by tanmay last updated on 20/Apr/19
$${leg}\:{A}_{\mathrm{1}} ,{A}_{\mathrm{2}} ,…{A}_{{n}} \:{and}\:{H}_{\mathrm{1}} ,{H}_{\mathrm{2}} ,…{H}_{{n}} \:{are}\:{n}\:{A}.{M}'{S}\: \\ $$$${and}\:{H}.{M}'{S}\:{respectively}\:{between}\:{a}\:{and}\:{b} \\ $$$${prove}\:{that}\:{A}_{{r}} {H}_{{n}−{r}+\mathrm{1}} ={ab} \\ $$$$\:{n}\geqslant{r}\geqslant\mathrm{1} \\ $$
Answered by Kunal12588 last updated on 20/Apr/19
$${a},{A}_{\mathrm{1}} ,{A}_{\mathrm{2}} ,…{A}_{{n}} ,{b} \\ $$$${are}\:{in}\:{AP} \\ $$$${where} \\ $$$${T}_{\mathrm{1}} ={a}, \\ $$$${T}_{{n}+\mathrm{2}} ={b} \\ $$$$\Rightarrow{T}_{\mathrm{1}} +\left({n}+\mathrm{1}\right){d}={b} \\ $$$$\Rightarrow{d}=\frac{{b}−{a}}{{n}+\mathrm{1}} \\ $$$${A}_{{r}} ={T}_{\mathrm{1}} +{rd} \\ $$$$\Rightarrow{A}_{{r}} ={a}+{r}\left(\frac{{b}−{a}}{{n}+\mathrm{1}}\right) \\ $$$$\Rightarrow{A}_{{r}} =\frac{{an}+{a}+{rb}−{ra}}{{n}+\mathrm{1}} \\ $$$$\Rightarrow{A}_{{r}} =\frac{{a}\left({n}−{r}+\mathrm{1}\right)+{rb}}{{n}+\mathrm{1}} \\ $$$${a},{H}_{\mathrm{1}} ,{H}_{\mathrm{2}} ,…{H}_{{n}} ,{b} \\ $$$${are}\:{in}\:{HP} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}},\frac{\mathrm{1}}{{H}_{\mathrm{1}} },\frac{\mathrm{1}}{{H}_{\mathrm{2}} },…,\frac{\mathrm{1}}{{H}_{{n}} },\frac{\mathrm{1}}{{b}}\:{are}\:{in}\:{AP} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{H}_{{r}} }=\frac{\frac{\mathrm{1}}{{a}}\left({n}−{r}+\mathrm{1}\right)+{r}\frac{\mathrm{1}}{{b}}}{{n}+\mathrm{1}}=\frac{{b}\left({n}−{r}+\mathrm{1}\right)+{ra}}{{ab}\left({n}+\mathrm{1}\right)} \\ $$$$\Rightarrow{H}_{{r}} =\frac{{ab}\left({n}+\mathrm{1}\right)}{{b}\left({n}−{r}+\mathrm{1}\right)+{ra}} \\ $$$${H}_{{n}−{r}+\mathrm{1}} =\frac{{ab}\left({n}+\mathrm{1}\right)}{{b}\left({n}−\left({n}−{r}+\mathrm{1}\right)+\mathrm{1}\right)+{a}\left({n}−{r}+\mathrm{1}\right)} \\ $$$$\Rightarrow{H}_{{n}−{r}+\mathrm{1}} =\frac{{ab}\left({n}+\mathrm{1}\right)}{{br}+{a}\left({n}−{r}+\mathrm{1}\right)} \\ $$$$\therefore{A}_{{r}} {H}_{{n}−{r}+\mathrm{1}} =\frac{{a}\left({n}−{r}+\mathrm{1}\right)+{rb}}{{n}+\mathrm{1}}×\frac{{ab}\left({n}+\mathrm{1}\right)}{{br}+{a}\left({n}−{r}+\mathrm{1}\right)} \\ $$$$\Rightarrow{A}_{{r}} {H}_{{n}−{r}+\mathrm{1}} ={ab}\left(\mathrm{1}\right) \\ $$$$\Rightarrow{A}_{{r}} {H}_{{n}−{r}+\mathrm{1}} ={ab} \\ $$
Commented by tanmay last updated on 20/Apr/19
$${excellent}…{i}\:{have}\:{posted}\:{these}\:{question}\left[{from}\right. \\ $$$${old}\:{book}\:{of}\:{BSTAT}\:{questions}… \\ $$
Commented by Kunal12588 last updated on 20/Apr/19
Sir I am in class 12. I got 70.8% in class 11, I know it's not good. What should I do after 12th. I am ok in maths, not so good in physics and very bad in chemistry.
Commented by tanmay last updated on 20/Apr/19
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