Differentiation (Part-38) | S N De

 

$\,1.\,$ Examine whether the following sequences are in H.P.

$\,(i)\,\,\{\frac 35,\frac{3}{11},\frac{3}{17},\frac{3}{23},\cdots\}$

Sol. We have to prove that $\,\{\frac 35,\frac{3}{11},\frac{3}{17},\frac{3}{23},\cdots\}\,$ are in H.P. 

For that we have to show that $\,\{\frac 53,\frac{11}{3},\frac{17}{3},\frac{23}{3},\cdots\}$ are in A.P.

Now, let $\,t_n\,$ denotes the $\,n\,$th term of the series.

So, $\,t_2-t_1\\=\frac{11}{3}-\frac{5}{3}\\=\frac{11-5}{3}\\=\frac{6}{3}\\=2$

Similarly,

$\,t_3-t_2\\=\frac{17}{3}-\frac{11}{3}\\=\frac{17-11}{3}\\=\frac 63\\=2$

Again, $\,t_4-t_3\\=\frac{23}{3}-\frac{17}{3}\\=\frac{23-17}{3}\\=\frac 63\\=2$

Hence, we notice that

$\,t_2-t_1=t_3-t_2=t_4-t_3=\cdots=2$

Hence,  $\,\{\frac 53,\frac{11}{3},\frac{17}{3},\frac{23}{3},\cdots\}$ are in A.P.

and so $\,\{\frac 35,\frac{3}{11},\frac{3}{17},\frac{3}{23},\cdots\}\,$ are in H.P. 

$\,1.\,$ Examine whether the following sequences are in H.P.

$\,(ii)\,\,\{\frac 54,\frac{10}{13},\frac{10}{18},\frac{10}{23},\cdots\}$

Sol. We have to prove that $\,\,\{\frac 54,\frac{10}{13},\frac{10}{18},\frac{10}{23},\cdots\}\,$ are in H.P. 

For that we have to show that $\,\{\frac 45,\frac{13}{10},\frac{18}{10},\frac{23}{10},\cdots\}$ are in A.P.

Now, let $\,t_n\,$ denotes the $\,n\,$th term of the series.

So, $\,t_2-t_1\\=\frac{13}{10}-\frac{4}{5}\\=\frac{13-8}{10}\\=\frac{5}{10}\\=\frac 12$

Similarly,

$\,t_3-t_2\\=\frac{18}{10}-\frac{13}{10}\\=\frac{18-13}{10}\\=\frac{5}{10}\\=\frac 12$

Again, $\,t_4-t_3\\=\frac{23}{10}-\frac{18}{10}\\=\frac{23-18}{10}\\=\frac{5}{10}\\=\frac 12$

Hence, we notice that

$\,t_2-t_1=t_3-t_2=t_4-t_3=\cdots=\frac 12$

Hence, $\,\{\frac 45,\frac{13}{10},\frac{18}{10},\frac{23}{10},\cdots\}$ are in A.P.

and so $\,\,\{\frac 54,\frac{10}{13},\frac{10}{18},\frac{10}{23},\cdots\}\,$ are in H.P. 

$\,1.\,$ Examine whether the following sequences are in H.P.

$\,(iii)\,\,\{\frac 17,\frac{1}{9},\frac{1}{12},\frac{1}{16},\cdots\}$

Sol. We have to prove that $\,\,\,\{\frac 17,\frac{1}{9},\frac{1}{12},\frac{1}{16},\cdots\}$ are in H.P. 

For that we have to show that $\,\{7,9,12,16,\cdots \}$ are in A.P.

Now, let $\,t_n\,$ denotes the $\,n\,$th term of the series.

So, $\,t_2-t_1\\=9-7\\=2$

Similarly,

$\,t_3-t_2\\=12-9\\=3$

Again, $\,t_4-t_3\\=16-12\\=4$

Hence, we notice that

$\,t_2-t_1 \neq t_3-t_2\neq t_4-t_3$

Hence, $\,\{7,9,12,16,\cdots \}$  are not in A.P.

and so $\,\,\,\{\frac 17,\frac{1}{9},\frac{1}{12},\frac{1}{16},\cdots\}$  are not in H.P. 

To download full PDF solution of Sequence and Series(Part-3), class XI for Infinite G.P. and H.P., click here . 

$\,1.\,$ Examine whether the following sequences are in H.P.

$\,(iv)\,\,\{\frac{1}{a-3b},\frac{1}{a-2b},\frac{1}{a-b},\frac{1}{a},\cdots\}$

Sol. We have to prove that $\,\,\,\{\frac{1}{a-3b},\frac{1}{a-2b},\frac{1}{a-b},\frac{1}{a},\cdots\}\,$ are in H.P. 

For that we have to show that $\,\{a-3b,a-2b,a-b,a,\cdots \}$ are in A.P.

Now, let $\,t_n\,$ denotes the $\,n\,$th term of the series.

So, $\,t_2-t_1\\=(a-2b)-(a-3b)\\=b$

Similarly,

$\,t_3-t_2\\=(a-b)-(a-2b)\\=b$

Again, $\,t_4-t_3\\=a-(a-b)\\=b$

Hence, we notice that

$\,t_2-t_1 = t_3-t_2= t_4-t_3=\cdots=b$

Hence, $\,\{a-3b,a-2b,a-b,a,\cdots \}$  are  in A.P.

and so $\,\,\,\{\frac{1}{a-3b},\frac{1}{a-2b},\frac{1}{a-b},\frac{1}{a},\cdots\}\,$  are  in H.P. 

$\,1.\,$ Examine whether the following sequences are in H.P.

$\,(v)\,\,\{\frac{1}{c+3d},\frac{1}{c+2d},\frac{1}{c},\frac{1}{c-3d},\cdots\}$

Sol. We have to prove that $\,\,\,\{\frac{1}{c+3d},\frac{1}{c+2d},\frac{1}{c},\frac{1}{c-3d},\cdots\}\,$ are in H.P. 

For that we have to show that $\,\{c+3d,c+2d,c,c-3d,\cdots \}$ are in A.P.

Now, let $\,t_n\,$ denotes the $\,n\,$th term of the series.

So, $\,t_2-t_1\\=(c+2d)-(c+3d)\\=-d$

Similarly,

$\,t_3-t_2\\=(c)-(c+2d)\\=-2d$

Again, $\,t_4-t_3\\=(c-3d)-c\\=-3d$

Hence, we notice that

$\,t_2-t_1 \neq t_3-t_2\neq  t_4-t_3$

Hence, $\,\{c+3d,c+2d,c,c-3d,\cdots \}$  are  not in A.P.

and so $\,\,\,\{\frac{1}{c+3d},\frac{1}{c+2d},\frac{1}{c},\frac{1}{c-3d},\cdots\}\,$   are not in H.P. 

$\,2(i).\,$ Find the H.P. whose first and second terms are $\,\frac 13\,$ and $\,\frac 15\,$ respectively.

Sol. To find the H.P. whose first two terms are $\,\frac 13\,$ and $\,\frac 15,\,$ we need to find A.P. whose first two terms are  $\,3\,$ and  $\,5.$

Let $\,t_n\,$ denotes the $\,n\,$th term of the A.P.

Suppose $\,t_1=3,\,\,t_2=5.$

So, $\,d=t_2-t_1=5-3=2,\\t_3=t_2+d=5+2=7,\\ t_4=t_3+d=7+2=9,\\ \vdots $

So, the A.P. is : $\,\{3,5,7,9,\cdots \}$ so that the required H.P. is $\,\{\frac 13,\frac 15, \frac 17,\frac 19,\cdots \}.$

$\,2(ii)\,$ If the $\,10\,$th and $\,19\,$th terms of a H.P. are $\,\left(-\frac{5}{37}\right)\,$ and $\,\left(-\frac{5}{64}\right)\,$ respectively, find the $\,n\,$th term of the H.P.

Sol. Let $~~\,u_{10}=-\frac{5}{37},\,u_{19}=-\frac{5}{64};$ where $\,u_{10},\,u_{19}\,$ are the $\,10\,$th and $\,19\,$th terms of a H.P.

Suppose that $\,t_{10}\,$ and $\,t_{19}\,$ are the $\,10\,$th and $\,19\,$th terms of the corresponding  A.P. and $\,a\,$ is the first term of the A.P.

Now, $\,t_{10}=-\frac{37}{5},\,\,t_{19}=-\frac{64}{5}\rightarrow(1)$

Now from $\,(1),\,$ we get, 

$\,a+(10-1)d=-\frac{37}{5} \Rightarrow a+9d=-\frac{37}{5} \rightarrow(2),\\~ a+(19-1)d=-\frac{64}{5} \Rightarrow a+18d=-\frac{64}{5}\rightarrow(3).$

Solving $\,(2),\,(3),\,$ we get,

$~~~~~(a+9d)-(a+18d)=-\frac{37}{5}+\frac{64}{5} \\ \Rightarrow -9d=\frac{27}{5} \\ \Rightarrow d=\frac{-3}{5}\rightarrow(4).$

Solving $\,(2),(4),\,$ we get,

$\,a+9 \times \left(-\frac 35\right)=-\frac{37}{5}\\ \Rightarrow a=-\frac{37}{5}+\frac{27}{5} \\ \Rightarrow a=\frac{-37+27}{5} \\ \Rightarrow a=\frac{-10}{5}=-2.$

So, $\,t_n=a+(n-1)d\\ \Rightarrow t_n=-2+(n-1) \times \left(-\frac 35\right)\\~~~~~~~~=\frac{-10+3-3n}{5}\\~~~~~~~~=\frac{-(3n+7)}{5}\\ \therefore u_n=\frac{5}{-(3n+7)}\,\,\,\text{(ans.)}$

$\,2(iii)\,$ If the $\,p\,$th term of a H.P. be $\,q\,$ and the $\,q\,$th term be equal to $\,p,\,$ show that, its $\,(p+q)\,$th term is $\,\frac{pq}{p+q}.$

Sol. Let $\,u_n\,$ be the $\,n\,$the term of the H.P. and  $\,t_n\,$ be the $\,n\,$the term of the corresponding  A.P. Also, suppose that $\,a\,$ be the first term and $\,d\,$ be the common difference of the A.P. 

According to the problem, 

$\,~~~~u_p=q \\ \Rightarrow t_p=\frac 1q \\ \Rightarrow a+(p-1)d=\frac 1q\rightarrow(1)\\ ~~~~~u_q=p \\ \Rightarrow t_q=\frac 1p \\ \Rightarrow a+(q-1)d=\frac 1p\rightarrow(2)$

Solving $\,(1),\,(2)\,$ we get,

$\,[a+(p-1)d]-[a+(q-1)d]=\frac 1q-\frac 1p\\ \Rightarrow (p-q)d=\frac{p-q}{pq}\\ \Rightarrow d=\frac{1}{pq}\rightarrow(3)$

Soving $\,(1),\,(3)\,$ we get,

$\,a+(p-1)\times \frac{1}{pq}=\frac 1q \\ \Rightarrow a=\frac 1q-\frac{p-1}{pq} \\ \Rightarrow a=\frac{p-(p-1)}{pq}=\frac{1}{pq}.$

Hence, $\,t_{p+q}=a+(p+q-1)d\\~~~~~~~~=\frac{1}{pq}+\frac{p+q-1}{pq}\\~~~~~~~~=\frac{1+(p+q-1)}{pq}\\~~~~~~~~=\frac{p+q}{pq}\\ \Rightarrow u_{p+q}=\frac{pq}{p+q}\,\,\text{(ans.)}$