Differentiation (Part-38) | S N De

 

$\,11.\,$ A boy saves $\,1\,$ paisa the first day, $\,2\,$ paise the second day, $\,4\,$ paise the third day and $\,8\,$ paise the fourth day and so on. Find his total savings at the end of $\,10\,$ days.

Sol. Clearly, the savings by the boy form an G.P. where the first term $\,(a)\,=1\,$paisa and common ratio $\,(r)=\frac 21=1.$

Hence, his total savings at the end of $\,10\,$ days is 

$=a \cdot\frac{r^n-1}{r-1}=1 \cdot \frac{2^{10}-1}{2-1}=1024-1=1023.$

So, the total savings of the boy at the end of $\,10\,$ days is $\,1023\,$ paisa or $\,10.23\,$ rupees.

$\,12.\,$ Shri A. K. Roy borrows Rs. $\,6000\,$ to repay the amount with interest of Rs. $\,138\,$ in $\,10\,$ monthly instalments, each instalment being twice the preceding one. Find the second and last instalments.

Sol. Total amount to repay rs.$=(6000+138)=6138.$ 

Since each instalment being twice the preceding one, let the first instalment being $\,a,\,$ second being $\,2a,\,$ third instalment being $\,3a\,$ and so on.

Here, the series of instalments are in G.P. , with common ratio $(r)=2.$

So, the sum of G.P. upto $\,10\,$ terms $=a \cdot \frac{r^n-1}{r-1}\\=a \cdot \frac{2^{10}-1}{2-1}\,\,\,[\because n=10]\\=a \cdot(1024-1)\\=1023a$ 

According to the problem, $\,1023a=6138 \Rightarrow a =\frac{6138}{1023}=6.$

So, the second instalment is : $=2a=2 \times 6=12\,$ rupees.

Last Instalment is : $=a \cdot r^{n-1}= 6\cdot 2^{10-1}=6 \cdot 2^9=3072\,$ rupees.

$\,13.\,$ A man borrows Rs. $\,19682\,$ without interest and agrees to repay the money in $\,9\,$ monthly instalments, each instalment being thrice the preceding one. After the seventh instalment has been paid, he wants to repay the balance in lump. How much has he to pay now?

Sol. Let the first instalment be $\,a\,$ rupees. So, the other instalments are $\,\,a,3a,9a,27a,\cdots \text{to 9 terms}.$

According to the problem,

$\,a \cdot \frac{3^9-1}{3-1}=19682 \\ \Rightarrow a \cdot \frac 12(19683-1)=19682 \\ \Rightarrow \frac a2 \cdot 19682=19682 \\ \Rightarrow a=2.$

Now, after the $\,7\,$ th instalment, other $\,2\,$ instalments are :

$=2 \cdot 3^{8-1}+ 2 \cdot 3^{9-1}\\=2 \cdot (3^7+3^8)\\=2 \cdot (2187+6561)\\=2 \times 8748\\=17496.$

So, after the seventh instalment has been paid, if he wants to repay the balance in lump, he has to pay rs. $\,17496.$

$\,14.\,$ The A.M. of two numbers is $\,15\,$ and their G.M. is $\,12.\,$ Find the numbers.

Sol. Let the two numbers be $\,a,b.$

So, according to the problem,

$\,\frac{a+b}{2}=15 \rightarrow(1),\\ \sqrt{ab}=12 \rightarrow(2).$

From $\,(1),\,$ we get, $\,ab=144\rightarrow(3)$

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

$\,a(30-a)=144 \\ \Rightarrow 30a-a^2=144 \\ \Rightarrow a^2-30a+144=0\\ \Rightarrow a^2-24a-6a+144=0 \\ \Rightarrow a(a-24)-6(a-24)=0 \\ \Rightarrow (a-24)(a-6)=0 \\ \Rightarrow a=24,6.$

Now, if $\,a=6,\,b=\frac{144}{6}=24.\,\,[\text{By (3)}]$

Again, if $\,a=24,\,b=\frac{144}{24}=6.\,\,[\text{By (3)}]$

Hence, the numbers are $\,6,24.$

$\,15.\,$ If $\,a, b, c\,$ are in G.P. and $\,x, y\,$ be the A.M. between $\,a,b\,$ and $\,b,c\,$ respectively, prove that,

$\,(i)\, \frac ax+\frac cy=2~~~~(ii)\,\frac 1x+\frac 1y=\frac 2b.$

Sol.  $\,\because \,\, a,b,c\,$ are in G.P. , $\,\,b^2=ac \rightarrow(1)$

Again, $\,x=\frac 12(a+b),\,\,y=\frac 12(b+c).$

So, $\,(i)\,\, \frac ax+\frac cy=\frac{2a}{a+b}+\frac{2c}{b+c}\\ \Rightarrow \frac ax+\frac cy=\frac{2ab+2ac+2ac+2bc}{(a+b)(b+c)}\\ \Rightarrow \frac ax+\frac cy=\frac{2a(b+c)+2b^2+2bc}{(a+b)(b+c)}\\~~~~~~~~~~~~~~~~ =\frac{2a(b+c)+2b(b+c)}{(a+b)(b+c)}\\~~~~~~~~~~~~~~~~ =\frac{2(b+c)(a+b)}{(a+b)(b+c)}\\~~~~~~~~~~~~~~~~ =\frac{2(a+b)(b+c)}{(a+b)(b+c)}\\~~~~~~~~~~~~~~~~ =2\,\,\text{(proved)}$

Sol. $\,(ii)\,\frac 1x+\frac 1y=\frac{2}{a+b}+\frac{2}{b+c}\\ \Rightarrow \frac 1x+\frac 1y=\frac{2b+2c+2a+2b}{(a+b)(b+c)}\\~~~~~~~~~~~~~~~~~=\frac{2a+4b+2c}{ab+ac+b^2+bc}\\~~~~~~~~~~~~~~~~~=\frac{2(a+2b+c)}{ab+b^2+b^2+bc}\\~~~~~~~~~~~~~~~~~=\frac{2(a+2b+c)}{b(a+2b+c)}\\~~~~~~~~~~~~~~~~~=\frac 2b \\ \Rightarrow \frac 1x+\frac 1y=\frac 2b\,\,\text{(proved)}$

$\,16.\,$ If $\,a,b,c\,$ be in A.P. and $\,x,y,z\,$ be in G.P., prove that,

$\,x^{b-c}. y^{c-a}.z^{a-b}=1.$

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

$\therefore \, b-a=c-b \\~~~~~~~~~ \Rightarrow a-b=b-c\rightarrow(1)$

Again, since $\,x,y,z\,$ are in G.P., $\,y^2=xz\rightarrow(2)$

Now, $\,x^{b-c}. y^{c-a}.z^{a-b}\\~~~~~~~~~=x^{b-c}.z^{b-c}.y^{c-a}\,\,[\text{By (1)}]\\~~~~~~~~~=(xz)^{b-c}.y^{c-a}\\~~~~~~~~~=(y^2)^{b-c}.y^{c-a}\\~~~~~~~~~=y^{2b-2c+c-a}\\~~~~~~~~~=y^{2b-c-a}\\~~~~~~~~~=y^{2b-2b}\,\,[\because 2b=a+c]\\~~~~~~~~~=y^0\\~~~~~~~~~=1\,\,\text{(proved)}$

$\,17(i).\,$ The sum of three numbers in G.P. is $\,70.\,$ If each of the two extremes be multiplied by $\,4\,$ and the mean by $\,5,\,$ the products are in A.P. Find the numbers.

Sol. Let three numbers be $\,a,ar,ar^2\,\,(a \neq 0). $

$\,\,\therefore a+ar+ar^2=70\rightarrow(1)$

If each of the two extremes be multiplied by $\,4\,$ , then the numbers turn out to be  $\,4a,5ar,4ar^2\,\,$ which form an A.P.

$\,\therefore \frac{4a+4ar^2}{2}=5ar\\ \Rightarrow 2a+2ar^2=5ar\\ \Rightarrow 2+2r^2=5r\\ \Rightarrow 2r^2-5r+2=0\\ \Rightarrow 2r^2-4r-r+2=0 \\ \Rightarrow 2r(r-2)-1(r-2)=0\\ \Rightarrow (r-2)(2r-1)=0 \\ \Rightarrow r=2,\frac 12.$

Putting $\,r=2,\,$ in $\,(1),\,$ we get,

$\,a+2a+2^2a=70 \\ \Rightarrow 7a=70 \\ \Rightarrow a=\frac{70}{7}=10.$

Hence, the three numbers are $\,10,2 \times 10, 2^2 \times 10\\=10,20,40.$

Putting $\,r=\frac 12,\,$ in $\,(1),\,$ we get,

$\,a+\frac 12a+(\frac 12)^2a=70 \\ \Rightarrow \frac 74a=70 \\ \Rightarrow a=\frac{70\times 4}{7}=40.$

Hence, the three numbers are $\,40,\frac 12 \times 40, (\frac 12)^2 \times 40\\=40,20,10.$

$\,17(ii)\,\,$ Let $\,t_r\,$ be the $\,r\,$th term of an A.P. If $\,t_2,\,t_8\,$ and $\,t_{32}\,$are in G.P., then find the common ratio of the G.P.

Sol. Let $\,a\,$ be the first term of the A.P. and $\,d\,$ being the common ratio.

$\,\therefore \, t_2=a+(2-1)d=a+d,\\ t_8=a+(8-1)d=a+7d, \\ t_{32}=a+(32-1)d=a+31d.$

If $\,t_2,\,t_8\,$ and $\,t_{32}\,$are in G.P.,

$\,\frac{t_8}{t_2}=\frac{t_{32}}{t_8}\\ \Rightarrow \frac{a+7d}{a+d}=\frac{a+31d}{a+7d}=\frac{(a+31d)-(a+7d)}{(a+7d)-(a+d)}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{24d}{64d}=4\rightarrow(1)$

Hence, from $\,(1),\,$ we can say that the common ratio of the G.P. is  $\,(r)=\frac{a+7d}{a+d}=4.$