Differentiation (Part-38) | S N De

 $\,5.\,$ If $\,a,b,c\,$ are in H.P., prove that

$\,(i)\,~~\frac{a}{b+c-a},\frac{b}{c+a-b},\frac{c}{a+b-c}\,$ are in H.P.

Sol. We have to prove that 

$\,\,~~\frac{a}{b+c-a},\frac{b}{c+a-b},\frac{c}{a+b-c}\,$ are in H.P.

i.e., $\,\,~~\frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}\,$ are in A.P.

i.e., $\,\,~~\left(\frac{b+c-a}{a}+1\right),\left(\frac{c+a-b}{b}+1\right),\left(\frac{a+b-c}{c}+1\right)\,$ are in A.P.

i.e., $\,\,~~\left(\frac{b+c-a+a}{a}\right),\left(\frac{c+a-b+b}{b}\right),\left(\frac{a+b-c+c}{c}\right)\,$ are in A.P.

i.e., $\,\,~~\left(\frac{b+c}{a}\right),\left(\frac{c+a}{b}\right),\left(\frac{a+b}{c}\right)\,$ are in A.P.

i.e., $\,\,~~\left(\frac{b+c}{a}+1\right),\left(\frac{c+a}{b}+1\right),\left(\frac{a+b}{c}+1\right)\,$ are in A.P.

i.e., $\,\,~~\left(\frac{b+c+a}{a}\right),\left(\frac{c+a+b}{b}\right),\left(\frac{a+b+c}{c}\right)\,$ are in A.P.

i.e., $\,\,~~\frac{1}{a},\frac{1}{b},\frac{1}{c}\,$ are in A.P. $~~[\because (a+b+c)\neq 0]$

i.e., $\,a,\,b,\,c\,$ are in H.P. which is true.

Hence, completes the proof.

$\,5.\,$ If $\,a,b,c\,$ are in H.P., prove that

$\,(ii)\,~~\frac{1}{bc},\frac{1}{ca},\frac{1}{ab}\,$ are in H.P.

Sol. We have to prove that 

$\,\,~~\frac{1}{bc},\frac{1}{ca},\frac{1}{ab}\,$ are in H.P.

i.e., $\,\,~~bc,ca,ab\,$ are in A.P.

i.e., $\,\,\frac{bc}{abc},\frac{ca}{abc},\frac{ab}{abc}\,$ are in A.P. $\,\,[\because abc \neq 0]$

i.e., $\,\,~~\frac 1a,\frac 1b,\frac 1c\,$ are in A.P.

i.e., $\,a,\,b,\,c\,$ are in H.P. which is true.

Hence, completes the proof.

$\,5.\,$ If $\,a,b,c\,$ are in H.P., prove that

$\,(iii)\,a(b+c),\,b(c+a),\,c(a+b)\,$ are in A.P.

Sol. We have to prove that 

$\,a(b+c),\,b(c+a),\,c(a+b)\,$ are in A.P.

i.e., $\,\,\frac{a(b+c)}{abc},\frac{b(c+a)}{abc},\frac{c(a+b)}{abc}\,$ are in A.P. $\,[\because abc \neq 0]$

i.e., $\,\,\frac{b+c}{bc},\frac{c+a}{ca},\frac{a+b}{ab}\,$ are in A.P. 

i.e., $\,\,\frac 1b+\frac 1c,\,\frac 1c+\frac 1a,\,\frac 1a+\frac 1b\,\,\,$ are in A.P. 

Since, $\,a,b,c\,$ are in H.P., $\,\frac 1a,\frac 1b,\frac 1c\,$ are in A.P.

and so, $\,\frac 1b-\frac 1a=\frac 1c-\frac 1b \\ \Rightarrow \frac 1a-\frac 1b=\frac 1b-\frac 1c\, \rightarrow(1)$

Now, to prove $\,\,\frac 1b+\frac 1c,\,\frac 1c+\frac 1a,\,\frac 1a+\frac 1b\,\,\,$ are in A.P. , 

we notice $\,\left(\frac 1c+\frac 1a\right)-\left(\frac 1b+\frac 1c\right)\\=\frac 1a-\frac 1b \rightarrow(2)$

Again, $\,\left(\frac 1a+\frac 1b\right)-\left(\frac 1c+\frac 1a\right)\\=\frac 1b-\frac 1c \rightarrow(3)$

So, from $\,(1),(2),(3)\,$, we can conclude that

$\left(\frac 1c+\frac 1a\right)-\left(\frac 1b+\frac 1c\right)\\=\left(\frac 1a+\frac 1b\right)-\left(\frac 1c+\frac 1a\right)$

Hence,$\,\,\frac 1b+\frac 1c,\,\frac 1c+\frac 1a,\,\frac 1a+\frac 1b\,\,\,$ are in A.P. 

and so, $\,a(b+c),\,b(c+a),\,c(a+b)\,$ are in A.P. (proved)

$\,5(iv)\,\,\frac{a}{a-b}=\frac{a+c}{a-c}$

Sol. Since $\,a,b,c\,$ are in H.P. , $\,\frac 1a,\frac 1b,\frac 1c\,$ are in A.P.

So, $\,\,~~~\frac 2b=\frac 1a+\frac 1c \\ \Rightarrow \frac 1c=\frac 2b-\frac 1a \\ \Rightarrow \frac 1c=\frac{2a-b}{ab}\\ \Rightarrow c=\frac{ab}{2a-b}.$

Now, $\,\frac{a+c}{a-c}\\=\frac{a+\frac{ab}{2a-b}}{a-\frac{ab}{2a-b}}\\=\frac{a(2a-b)+ab}{a(2a-b)-ab}\\=\frac{a(2a-b+b)}{a(2a-b-b)}\\=\frac{2a}{2(a-b)}\\=\frac{a}{a-b}\,\,\text{(proved)}$

$\,6(i).\,$ If $\,a,b,c\,$ are in A.P., show that, $\,\frac{bc}{a(b+c)},\frac{ca}{b(c+a)}\,$ and $\,\frac{ab}{c(a+b)}\,$ are in H.P. 

Sol. Since $\,a,b,c\,$ are in A.P., 

$\,b-a=c-b=k,\,\text{(say)}.$

Now, we have to prove that

 $\,\frac{bc}{a(b+c)},\frac{ca}{b(c+a)}\,$ and $\,\frac{ab}{c(a+b)}\,$ are in H.P. 

i.e.,  $\,\frac{a(b+c)}{bc},\frac{b(c+a)}{ca}\,$ and $\,\frac{c(a+b)}{ab}\,$ are in A.P. 

i.e.,  $\,\frac{a^2(b+c)}{abc},\frac{b^2(c+a)}{abc}\,$ and $\,\frac{c^2(a+b)}{abc}\,$ are in A.P. 

i.e., $\,a^2(b+c),b^2(c+a),c^2(a+b)\,$ are in A.P. $\,[\because abc \neq 0]$

Now, $\,b^2(c+a)-a^2(b+c)\\=c(b^2-a^2)+ab(b-a)\\=(b-a)[c(b+a)+ab]\\=k[b(c+a)+ac]\\=k[\frac 12 (c+a)^2+ca]\rightarrow(1)$

Again, $c^2(a+b)-b^2(c+a)\\=a(c^2-b^2)+cb(c-b)\\=(c-b)[a(c+b)+cb]\\=k[b(c+a)+ca]\\=k[\frac 12(c+a)^2+ca] \rightarrow(2)$

Hence, from $\,(1),(2)\,$ we get, 

$b^2(c+a)-a^2(b+c)=c^2(a+b)-b^2(c+a)\,$ and so

$\,a^2(b+c),b^2(c+a),c^2(a+b)\,$ are in A.P. $\,[\because abc \neq 0]$ and hence

 $\,\frac{bc}{a(b+c)},\frac{ca}{b(c+a)}\,$ and $\,\frac{ab}{c(a+b)}\,$ are in H.P. (proved)

$\,6(ii)\,$ If $\,a^2,b^2,c^2\,$ are in A.P., show that, $\,(b+c),(c+a),(a+b)\,$ are in H.P.

Sol. Given that $\,a^2,b^2,c^2\,$ are in A.P. 

i.e., $\,(a^2+ab+bc+ca),(b^2+ab+bc+ca),\\~~~~(c^2+ab+bc+ca)\,$ 

are in A.P. 

i.e., $\,(a+b)(a+c),\,(b+c)(b+a),\,(c+a)(c+b)\,$ are in A.P.

Now, Dividing each term by $\,(a+b)(b+c)(c+a),\,$ we get,

$\,\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}\,$ are in A.P. (showed)

$\,6(iii)\,$ If four positive numbers $\,a,b,c\,$ and $\,d\,$ are in A.P. then prove that $\,abc,\,abd,\,acd\,$ and $\,bcd\,$ are in H.P.

Sol. Given $\,a,b,c,d\,$ are in A.P.

So $\,d,c,b,a\,$ are in A.P.

$\,\Rightarrow 1/d, 1/c, 1/b, 1/a\,$ are in H.P.

So $\,\frac{abcd}{d},\frac{abcd}{c}, \frac{abcd}{b},\frac{abcd}{a}\,$ are in H.P.

$\,\Rightarrow abc, abd, acd, bcd\,$ are in H.P.

$\,7(i)\,$ Prove that , $\,a,b,c\,$ are in A.P., G.P. or H.P. accordingly as $\,\frac{a-b}{b-c}=1\,$ or, $\,\frac ab\,$ or, $\,\frac ac.$

Sol. $\,\frac{a-b}{b-c}=1 \\ \Rightarrow a-b=b-c \\ \Rightarrow a+c=2b \\ \Rightarrow b=\frac{a+c}{2} \rightarrow(1)$

Hence, $\,(1)\,$ shows that $\,a,b,c\,$ are in A.P.

Again, $\,\frac{a-b}{b-c}=\frac ab \\ \Rightarrow b(a-b)=a(b-c) \\ \Rightarrow ab-b^2=ab-ac \\ \Rightarrow b^2=ac \rightarrow(2)$

Hence, $\,(2)\,$ shows that $\,a,b,c\,$ are in G.P.

Finally, $\,\frac{a-b}{b-c}=\frac ac \\ \Rightarrow c(a-b)=a(b-c) \\ \Rightarrow ac-bc=ab-ac \\ \Rightarrow 2ac=b(a+c) \\ \Rightarrow b=\frac{2ac}{a+c}\rightarrow(3)$  

Hence, $\,(3)\,$ shows that $\,a,b,c\,$ are in H.P.

$\,7(ii)\,$ If $\,a^x=b^y=c^z\,\,$ and $\,a,b,c\,$ are in G.P., prove that, $\,x,y,z\,$ are in H.P.

Sol. Since $\,a,b,c\,$ are in G.P.,

$\,b^2=ac \\ \Rightarrow \log (b^2)=\log(ac) \\ \Rightarrow 2\log b=\log a+\log c\rightarrow(1)$

Since $\,a^x=b^y=c^z \\ \Rightarrow \log(a^x)=\log(b^y)=\log(c^z) \\ \Rightarrow x\log a=y\log b\\~~~~~=z \log c=k(\neq 0)\,\text{(say)}\rightarrow(2) $ 

Hence, from $\,(1),\,(2)\,$ we get,

$\,2 \cdot\frac ky=\frac kx+\frac kz \\ \Rightarrow \frac 2y=\frac 1x+\frac 1z \rightarrow(3)$

Hence, from $\,(3),\,$ we can conclude that $\,x,y,z\,$ are in H.P.

$\,7(iii)\,$ If $\,p,q,r\,$ are in A.P., $\,q,r,s\,$ are in G.P. and $\,r,s,t\,$ are in H.P., prove that $\,p,r,t\,$ are in G.P.

Sol. If $\,p,q,r\,$ are in A.P., 

$\,2q=p+r \Rightarrow q=\frac{p+r}{2}\rightarrow(1)$

If  $\,q,r,s\,$ are in G.P., $\,r^2=qs\rightarrow(2)$

If $\,r,s,t\,$ are in H.P., $\,s=\frac{2rt}{r+t}\rightarrow(3)$

Hence, from $\,(1),(2),(3)\,$ we get,

$\,r^2=qs=\frac{p+r}{2} \times \frac{2rt}{r+t} \\ \Rightarrow r^2(r+t)=rt(p+r) \\ \Rightarrow 

 r(r+t)=t(p+r)\\ \Rightarrow  r^2+rt=pt+rt \\ \Rightarrow  r^2=pt \rightarrow(4)$

Hence, by $\,(4),\,$ we can conclude that  $\,p,r,t\,$ are in G.P.