Differentiation (Part-38) | S N De

 

$\,7.\,$ If the $\,n\,$th term of the series be $\,(2 \cdot 3^{n-1}+1),\,$ find the sum of first $\,r\,$ terms of the series. 

Sol. The sum of first $\,r\,$ terms of the series : 

$\,S_r=t_1+t_2+t_3+\cdots +t_r\\=(2 \cdot 3^{1-1}+1)+(2 \cdot 3^{2-1}+1)\\+(2 \cdot 3^{3-1}+1)+\cdots+(2 \cdot 3^{r-1}+1)\\=(2+1)+(2 \cdot 3+1)+(2 \cdot 3^2+1)\\+\cdots+(2 \cdot 3^{r-1}+1)\\=2(1+3+3^2+\cdots+3^{r-1})\\+(1+1+1+\cdots\text{to r terms})\\=2 \cdot \frac{3^r-1}{3-1}+r\\=3^r-1+r\,\,\text{(ans.)}$ 

$\,8.\,$ If $\,a,b,c\,$ be respectively the $\,p\,$th , $\,q\,$th and $\,r\,$th terms of a G.P. , prove that, $\,\,a^{q-r}\cdot b^{r-p} \cdot c^{p-q}=1.$

Sol. Let the first term of the G.P. be $\,l\,$ and common ratio being $\,m.$ and the $\,n\,$th term of the G.P. be denoted by $\,t_n.$

$\,\therefore \,a=lm^{p-1}=t_p,\\ ~~~~~~b=lm^{q-1}=t_q,\\~~~~~~c=lm^{r-1}=t_r.$

Now, $\,\,a^{q-r}\cdot b^{r-p} \cdot c^{p-q}\\=(lm^{p-1})^{q-r}\cdot (lm^{q-1})^{r-p} \cdot (lm^{r-1})^{p-q}\\=l^{q-r+r-p+p-q} \times m^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}\\=l^0 \times m^{p(q-r)+q(r-p)+r(p-q)} \\~~~~~~~~~~\times m^{-(q-r)-(r-p)-(p-q)}\\=l^0 \times m^0 \times m^0\\=1 \,\,\,\,\text{(proved)}$

$\,9.\,$ If $\,a,b,c\,$ and $\,d\,$ are in G.P., show that :

$\,(i)\,~~(a-b)^2,\,(b-c)^2,\,(c-d)^2\,\,$ are in G.P.

Sol. Let the common ratio of the G.P. be $\,r\,$ so that $\,\,b=ar,\,c=ar^2,\,\,d=ar^3.$

$\,\,\frac{(b-c)^2}{(a-b)^2}\\=\frac{(ar-ar^2)^2}{(a-ar)^2}\\=\frac{a^2r^2(1-r)^2}{a^2(1-r)^2}\\=r^2 \rightarrow(1)\\ \frac{(c-d)^2}{(b-c)^2}\\=\frac{(ar^2-ar^3)^2}{(ar-ar^2)^2}\\=\frac{a^2r^4(1-r)^2}{a^2r^2(1-r)^2}\\=r^2 \rightarrow(2)$

Hence, from $\,(1)\,$ and $\,(2)\,$ it follows that

$\,\,\frac{(b-c)^2}{(a-b)^2}=\frac{(c-d)^2}{(b-c)^2}$

and so $\,\,(a-b)^2,\,(b-c)^2,\,(c-d)^2\,\,$ are in G.P.

$\,9.\,$ If $\,a,b,c\,$ and $\,d\,$ are in G.P., show that :

$\,(ii)\,~~a^2+b^2,\,b^2+c^2,c^2+d^2\,\,\,$ are in G.P.

Sol. Let the common ratio of the G.P. be $\,r\,$ so that $\,\,b=ar,\,c=ar^2,\,\,d=ar^3.$

$\,\,~~~\frac{b^2+c^2}{a^2+b^2}\\=\frac{a^2r^2+a^2r^4}{a^2+a^2r^2}\\=\frac{a^2r^2(1+r^2)}{a^2(1+r^2)}\\=r^2 \rightarrow(1)\\~~~~~ \frac{c^2+d^2}{b^2+c^2}\\=\frac{a^2r^4+a^2r^6}{a^2r^2+a^2r^4}\\=\frac{a^2r^4(1+r^2)}{a^2r^2(1+r^2)}\\=r^2 \rightarrow(2)$

Hence, from $\,(1)\,$ and $\,(2)\,$ it follows that

$\,\,\frac{b^2+c^2}{a^2+b^2}=\frac{c^2+d^2}{b^2+c^2}$

and so $\,\,a^2+b^2,\,b^2+c^2,c^2+d^2\,\,$ are in G.P.

$\,9.\,$ If $\,a,b,c\,$ and $\,d\,$ are in G.P., show that :

$\,(iii)\,~~a^2+b^2+c^2,\,ab+bc+cd,b^2+c^2+d^2\,\,\,$ are in G.P.

Sol. Let the common ratio of the G.P. be $\,r\,$ so that $\,\,b=ar,\,c=ar^2,\,\,d=ar^3.$

$\,\,~~~\frac{ab+bc+cd}{a^2+b^2+c^2}\\=\frac{a^2r+a^2r^3+a^2r^5}{a^2+a^2r^2+a^2r^4}\\=\frac{a^2r(1+r^2+r^4)}{a^2(1+r^2+r^4)}\\=r \rightarrow(1)\\~~~~~ \frac{b^2+c^2+d^2}{ab+bc+cd}\\=\frac{a^2r^2+a^2r^4+a^2r^6}{a^2r+a^2r^3+a^2r^5}\\=\frac{a^2r^2(1+r^2+r^4)}{a^2r(1+r^2+r^4)}\\=r \rightarrow(2)$

Hence, from $\,(1)\,$ and $\,(2)\,$ it follows that

$\,\,\frac{ab+bc+cd}{a^2+b^2+c^2}=\frac{b^2+c^2+d^2}{ab+bc+cd}$

and so $\,\,a^2+b^2+c^2,\,ab+bc+cd,b^2+c^2+d^2\,\,$ are in G.P.

$\,9.\,$ If $\,a,b,c\,$ and $\,d\,$ are in G.P., show that :

$\,(iv)\,~~(b-c)^2+(c-a)^2+(d-b)^2=(a-d)^2.$ 

Sol. Let the common ratio of the G.P. be $\,r\,$ so that $\,\,b=ar,\,c=ar^2,\,\,d=ar^3.$

$\,(b-c)^2+(c-a)^2+(d-b)^2\\=(ar-ar^2)^2+(ar^2-a)^2+(ar^3-ar)^2\\=a^2r^2(1-r)^2+a^2(r^2-1)^2+a^2r^2(r^2-1)^2\\=a^2r^2(r-1)^2+a^2(r+1)^2(r-1)^2\\+a^2r^2(r+1)^2(r-1)^2\\=a^2(r-1)^2[r^2+(r+1)^2+r^2(r+1)^2]\\=a^2(r-1)^2[r^2+(r+1)^2+r^2(r^2+2r+1)]\\=a^2(r-1)^2[(r^2)^2+2r^2(r+1)+(r+1)^2]\\=a^2(r-1)^2(r^2+r+1)^2\\=a^2[(r-1)(r^2+r+1)]^2\\=a^2(r^3-1)^2\\=(ar^3-a)^2\\=(d-a)^2\\=(a-d)^2\,\,\,\text{(proved)}$

$\,10(i)\,\,$ If three numbers $\,\,a,b,c\,\,$ satisfy the relation $\,\,(a^2+b^2)(b^2+c^2)=(ab+bc)^2,\,\,$ show that $\,\,a,b,c\,$ are in G.P.

Sol. $\,\,(a^2+b^2)(b^2+c^2)=(ab+bc)^2\\ \Rightarrow a^2b^2+a^2c^2+b^4+b^2c^2=a^2b^2+2ab^2c+b^2c^2 \\ \Rightarrow a^2c^2+b^4=2ab^2c \\ \Rightarrow a^2c^2-2ab^2c+b^4=0 \\ \Rightarrow  (ac)^2-2(ac)b^2+(b^2)^2=0 \\ \Rightarrow  (ac-b^2)^2=0 \\ \Rightarrow  ac-b^2=0 \\ \Rightarrow  b^2=ac \rightarrow(1)$

Hence, from $\,(1),\,$ it follows that  $\,\,~a,b,c~\,$ are in G.P.

$\,10(ii)\,\,$ If $\,a,b,c\,$ are in geometric progression, then show that, 

$\,\,\frac{a^2+ab+b^2}{ab+bc+ca}=\frac{a+b}{b+c}$

Sol. $\,\,\because \,a,b,c\,$are in geometric progression,

$\,\,b^2=ac$

So, $\,\text{LHS}=\frac{a^2+ab+b^2}{ab+bc+ca}\\~~~~~~~~=\frac{a^2+ab+ac}{bc+b^2+ab}~~~~[\because b^2=ac]\\~~~~~~~~=\frac{a(a+b+c)}{b(c+b+a)}\\~~~~~~~~=\frac ab$

$\,\text{RHS}=\frac{a+b}{b+c}\\~~~~~~~~=\frac ab \cdot\frac{1+(b/a)}{1+(c/b)}\\~~~~~~~~=\frac ab \cdot\frac{1+(b/a)}{1+(b/a)}~~~[\because b^2=ac \Rightarrow \frac ba=\frac cb]\\~~~~~~~~= \frac ab$

Hence, L.H.S.=R.H.S. (showed)