Differentiation (Part-38) | S N De

 

$\,5(i)\,$ Find three numbers in G.P. whose sum is $\,19\,$ and product is $\,216\,$.

Sol.  Let the three numbers in G.P. be  $\,\frac ar,a,ar\,$ so that 

$\,\frac ar+a+ar=19 \rightarrow(1), \\ \frac ar \cdot a \cdot ar=216 \\ \Rightarrow a^3=216 \\ \Rightarrow a=\sqrt[3]{216}=6\rightarrow(2)$

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

$\,\frac 6r+6+6r=19 \\ \Rightarrow 6(r+\frac 1r)=19-6 \\ \Rightarrow 6 \times \frac{r^2+1}{r}=13 \\ \Rightarrow 6(r^2+1)=13r \\ \Rightarrow 6r^2-13r+6=0 \\ \Rightarrow 6r^2-4r-9r+6=0 \\ \Rightarrow 2r(3r-2)-3(3r-2)=0 \\ \Rightarrow (3r-2)(2r-3)=0 \\ \Rightarrow r=\frac 23,\, \frac 32.$

So, for $\,\,a=6,\, r=\frac 23,\,$ we get three numbers in G.P. as follows :

$\, \frac ar,a,ar\\= \frac{6}{2/3},\,6,\,6 \cdot \frac 23\\=9,6,4\,\,\text{(ans.)}$

Again, for $\,\,a=6,\, r=\frac 32,\,$ we get three numbers in G.P. as follows :

$\, \frac ar,a,ar\\= \frac{6}{3/2},\,6,\,6 \cdot \frac 32\\=4,6,9\,\,\text{(ans.)}$

$\,5(ii)\,$  Three rational numbers $\,x, y, z\,$ are in G.P. Sum of the numbers is $\,65\,$ and the product of the first and third numbers is $\,225\,$. Find the common ratio of the G.P.

Sol.  Let the three numbers in G.P. be $\,a, ar,ar^2\,$ so that 

$\,a+ar+ar^2=65 \rightarrow(1), \\ a \cdot ar^2=225 \\ \Rightarrow a^2r^2=225 \\ \Rightarrow ar=\pm\sqrt{225}=\pm 15\rightarrow(2)$

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

$\,\,a+15+ar^2=65 \\ \Rightarrow a(1+r^2)=65-15 \\ \Rightarrow  a(1+r^2)=50 \rightarrow(3)$

So, from $\,(1),\,(3)\,$ we get,

$\,\frac{a(1+r+r^2)}{a(1+r^2)}=\frac{65}{50} \\ \Rightarrow  \frac{1+r+r^2}{1+r^2}=\frac{13}{10} \\ \Rightarrow  1+\frac{r}{1+r^2}=\frac{13}{10} \\ \Rightarrow  \frac{r}{1+r^2}=\frac{13}{10}-1 \\ \Rightarrow  \frac{r}{1+r^2}=\frac{3}{10} \\ \Rightarrow 3(1+r^2)=10r \\ \Rightarrow  3r^2-10r+3=0 \\ \Rightarrow  3r^2-9r-r+3=0 \\ \Rightarrow  3r(r-3)-1(r-3)=0 \\ \Rightarrow (r-3)(3r-1)=0 \\ \Rightarrow  r=3,\frac 13.$ 

Hence, the common ratio of the G.P. is $\,3\,$ or $\,\,\frac 13\,\,\text{(ans.)}$

$\,5(iii)\,\,$ Find the three terms in G.P. whose product is $\,729\,$ and the sum of their product in pairs is $\,351.$

Sol. Let the three numbers in G.P. be  $\,\frac ar,a,ar\,$ so that 

$\,\frac ar \cdot a \cdot ar=729 \rightarrow(1), \\ \Rightarrow a^3=729\\ \Rightarrow a=\sqrt[3]{729}=9\rightarrow(2)$

Again, $\,\frac ar\cdot a+a\cdot ar+ar \cdot \frac ar=351 \\ \Rightarrow a^2(\frac 1r+r+1)=351 \\ \Rightarrow 9^2(\frac{1+r^2+r}{r})=351 \\ \Rightarrow 81(1+r^2+r)=351r \\ \Rightarrow 81r^2+81r-351r+81=0 \\ \Rightarrow 81r^2-270r+81=0 \\ \Rightarrow 27(3r^2-10r+3)=0 \\ \Rightarrow 3r^2-10r+3=0 \\ \Rightarrow 3r^2-9r-r+3=0 \\ \Rightarrow3r(r-3)-1(r-3)=0 \\ \Rightarrow (r-3)(3r-1)=0 \\ \Rightarrow r=3,\frac 13. $ 

Now, for $\,a=9, r=3,\,\,$ the three numbers in G.P. are :

$\,\frac ar, a , ar \\=\frac{9}{3},9, 9 \cdot 3\\=3,9,27\,\,\text{(ans.)}$

Again, for $\,a=9, r=\frac 13,\,\,$ the three numbers in G.P. are :

$\,\frac ar, a , ar \\=\frac{9}{1/3},9, 9 \cdot \frac 13\\=27,9,3\,\,\text{(ans.)}$

$\,5(iv)\,\,$ The sum of three consecutive numbers in G.P. is $\,21\,$ and the sum of their squares is $\,189.\,$ Find the numbers.

Sol.  Let the three numbers in G.P. be  $\,a,ar,ar^2\,$ so that 

$\,a+ar+ar^2=21 \\ \Rightarrow a(1+r+r^2)=21 \rightarrow(1) $

and $\,a^2+(ar)^2+(ar^2)^2=189 \\ \Rightarrow a^2(1+r^2+r^4)=189 \rightarrow(2)$

Now, from $\,(1),\,$ we get, 

$\,a^2(1+r+r^2)^2=21^2 \\ \Rightarrow a^2[(1+r)^2+2(1+r)r^2+r^4]=441 \\ \Rightarrow a^2[1+2r+r^2+2r^2+2r^3+r^4]=441 \\ \Rightarrow a^2[(1+r^2+r^4)+2r(1+r+r^2)]=441 \\ \Rightarrow a^2(1+r^2+r^4)\\+2(ra)a(1+r+r^2)=441 \\ \Rightarrow 189+2(ra)\times 21=441 \,\,[\text{By (1),(2)}]  \\ \Rightarrow ra=\frac{441-189}{2 \times 21}=6 \\ \Rightarrow a=\frac 6r\rightarrow(3)$

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

$\,\frac 6r(1+r+r^2)=21 \\ \Rightarrow \frac 6r+6+6r=21 \\ \Rightarrow 6 \times \frac{1+r+r^2}{r}=21 \\ \Rightarrow 6r^2+6r-21r+6=0  \\ \Rightarrow 6r^2-15r+6=0  \\ \Rightarrow  3(2r^2-5r+2)=0  \\ \Rightarrow  2r^2-5r+2=0 \\ \Rightarrow 2r^2-4r-r+2=0 \\ \Rightarrow 2r(r-2)-1(r-2)=0  \\ \Rightarrow  (r-2)(2r-1)=0  \\ \Rightarrow  r=2,\frac 12\rightarrow(4)$

Hence, for $\,r=2,\,a=\frac 6r=\frac{6}{2}=3.$

So, the three numbers in G.P. are  

$\,a,ar,ar^2\\=3, (3\cdot 2),(3 \cdot 2^2)\\=3,6,12\,\,\text{(ans.)}$ 

Again,  for $\,r=\frac 12,\,a=\frac 6r=\frac{6}{1/2}=12.$

So, the three numbers in G.P. are  

$\,a,ar,ar^2\\=12, (12\cdot \frac12),(12 \cdot \frac{1}{2^2})\\=12,6,3.\,\,\text{(ans.)}$ 

$\,6.\,$ Find the sum to $\,n\,$ terms of the following series :

$\,(i)\, (1)+(1+3)+(1+3+3^2)+\cdots$

Sol. Any term $\,t_k\,\,\,$of the given series is given by :

$\,t_k=1+3+3^2+\cdots+3^{k-1}\\~~~~=\frac{3^k-1}{3-1}\\~~~~=\frac 12(3^k-1)$

Hence, $\,S_n=(1)+(1+3)+(1+3+3^2)\\~~~+\cdots \text{to n terms}\\=t_1+t_2+t_3+\cdots +t_n\\=\frac 12(3-1)+\frac 12(3^2-1)+\frac 12(3^3-1)+\\ \cdots+\frac 12(3^n-1)\\=\frac 12(3+3^2+3^3+\cdots +3^n)\\~~~~-\frac 12(1+1+1+\cdots \text{to n terms})\\=\frac 12 \cdot 3 \cdot \frac{3^n-1}{3-1}-\frac n2\\=\frac 34 (3^n-1)-\frac n2\,\,\text{(ans.)}$

$\,6.\,$ Find the sum to $\,n\,$ terms of the following series :

$\,(ii)\, 1+3+7+15+31+\cdots$

Sol.  $\,S_n=1+3+7+15+31+\cdots \text{to n terms}\\=(2^1-1)+(2^2-1)+(2^3-1)+(2^4-1)\\~~~~+(2^5-1)+\cdots+(2^n-1)\\=(2+2^2+2^3+2^4+2^5+\cdots+2^n)\\-(1+1+1+1+1+\cdots \text{to n terms})\\=2 \cdot \frac{2^n-1}{2-1}-n\\=2(2^n-1)-n\\=2^{n+1}-2-n\\=2^{n+2}-(n+2) \,\,\text{(ans.)}$

$\,6.\,$ Find the sum to $\,n\,$ terms of the following series :

$\,(iii)\, 3+7+14+27+52+\cdots$

Sol.  $\,S_n=3+7+14+27+52+\cdots \text{to n terms}\\=3+(3 \times 2+1)+(3 \times 2^2+2)\\+(3 \times 2^3+3)+(3 \times 2^4+4)\\~~~~+\cdots+\{3 \times 2^{n-1}+(n-1)\}\\=(3+3 \times 2+3 \times  2^2+3 \times 2^3\\~~+\cdots+3 \times 2^{n-1})\\+(1+2+3+\cdots \text{to (n-1) terms})\\=3 \cdot \frac{2^n-1}{2-1}+\frac{(n-1)(n-1+1)}{2}\\=3(2^n-1)+\frac n2(n-1)\,\,\text{(ans.)}$