Maths NCERT Exemplar Solutions Class 11th Chapter Nine: Overview, Questions, Preparation

1. The first term of an A.P. is $a$ , and the sum of the first $p$ terms is zero, show that the

sum of its next $q$ terms is $\frac{-a\left(p+q\right)q}{p-1}$ .

[Hint: Required sum = $S_{p+q}-S_p$ ]

Sol:

$\begin{array}{l}Giventhat{a}_1=aand{S}_p\\ SumofnextqtermsofthegivenA.P.=S_{p+q}-S_p\\ \therefore S_{p+q}=\frac{p+q}{2}\left[2a+\left(p+q-1\right)d\right]\\ and{S}_p=\frac{p}{2}\left[2a+\left(p-1\right)d\right]=0\\ \Rightarrow 2a+\left(p-1\right)d=0\Rightarrow \left(p-1\right)d=-2a\\ \Rightarrow d=\frac{-2a}{p-1}\\ Sumofnextqterms=S_{p+q}-S_p\\ =\frac{p+q}{2}\left[2a+\left(p+q-1\right)d\right]-0\\ =\frac{p+q}{2}\left[2a+\left(p+q-1\right)\left(\frac{-2a}{p-1}\right)\right]\\ =\frac{p+q}{2}\left[2a+\frac{\left(p-1\right)\left(-2a\right)}{p-1}-\frac{2aq}{p-1}\right]\\ =\frac{p+q}{2}\left[2a-2a-\frac{2aq}{p-1}\right]=\frac{p+q}{2}\left(-\frac{2aq}{p-1}\right)\\ =\frac{-a\left(p+q\right)q}{p-1}\\ Hence,therequiredsum=\frac{-a\left(p+q\right)q}{p-1}\end{array}$

2. A man saved Rs 66000 in 20 years. In each succeeding year after the first year, he saved Rs 200 more than what he saved in the previous year. How much did he save in the first year?

Sol:

$\begin{array}{l}Letxbesavedinfirstyear.\\ Annualincrement=200\\ whichformsanA.P.\\ firstterm=aandcommondifferenced=200\\ n=20years\\ \therefore S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]=S_{20}=\frac{20}{2}\left[2a+\left(20-1\right)200\right]\\ \Rightarrow 66000=10\left[2a+3800\right]\Rightarrow 6600=2a+3800\\ \Rightarrow 2a=6600-3800\Rightarrow 2a=2800\Rightarrow a=1400\\ Hence,themansaved1400inthefirstyear.\end{array}$

1. If $A$ is the arithmetic mean and $G_1,G_2$ be two geometric means between any two numbers, then prove that

$2A=\frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}$

Sol:

$\begin{array}{l}Letthetwonumbersbexandy\\ \therefore A=\frac{x+y}{2}\dots \left(i\right)\\ If{G}_{1}and{G}_{2}bethegeometricmeansbetweenxandythenx,G_1,G_2,yareinG.P.\\ theny=x{r}^{4-1}\left[\because a_n=a{r}^{n-1}\right]\\ \Rightarrow y=x{r}^3\Rightarrow \frac{y}{x}=r^3\\ \Rightarrow r={\left(\frac{y}{x}\right)}^{1/3}\\ Now{G}_1=xr=x{\left(\frac{y}{x}\right)}^{1/3}\left[\because r={\left(\frac{y}{x}\right)}^{1/3}\right]\\ and{G}_2=x{r}^2=x{\left(\frac{y}{x}\right)}^{2/3}\\ \therefore FromRHS\frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}=\frac{x^2{\left(\frac{y}{x}\right)}^{2/3}}{x{\left(\frac{y}{x}\right)}^{2/3}}+\frac{x^2{\left(\frac{y}{x}\right)}^{4/3}}{x{\left(\frac{y}{x}\right)}^{1/3}}\\ =x+x{\left(\frac{y}{x}\right)}^{\frac{4}{3}-\frac{1}{3}}=x+x\left(\frac{y}{x}\right)=x+y=2ALHS\left[usingeq.\left(i\right)\right]\\ LHS=RHSHenceproved.\end{array}$

2. If ${\theta }_1,{\theta }_2,{\theta }_3,...,{\theta }_n$ are in A.P., whose common difference is $d$ , show that $sec{\theta }_{1}sec{\theta }_2+sec{\theta }_{2}sec{\theta }_3+...+sec{\theta }_{n-1}sec{\theta }_n=\frac{tan{\theta }_n-tan{\theta }_1}{sind}.$

Sol:

$\begin{array}{l}Since{\theta }_1,{\theta }_2,{\theta }_3,\dots ,{\theta }_{n}areinA.P.\\ \therefore {\theta }_2-{\theta }_1={\theta }_3-{\theta }_2=\dots ={\theta }_n-{\theta }_{n-1}=d\\ Nowwehavetoprovethat\\ sec{\theta }_1.sec{\theta }_2+sec{\theta }_2.sec{\theta }_3+\dots +sec{\theta }_{n-1}.sec{\theta }_n=\frac{tan{\theta }_n-tan{\theta }_1}{sind}\\ LHS.\frac{sind}{sind}\left[sec{\theta }_1.sec{\theta }_2+sec{\theta }_2.sec{\theta }_3+\dots +sec{\theta }_{n-1}.sec{\theta }_n\right]\\ Takingonly\frac{sind\left[sec{\theta }_1.sec{\theta }_2\right]}{sind}=\frac{sind\left[\frac{1}{cos{\theta }_1}.\frac{1}{cos{\theta }_2}\right]}{sind}\\ =\frac{sin\left({\theta }_2-{\theta }_1\right)}{sind}.\frac{1}{cos{\theta }_{1}cos{\theta }_2}\\ =\frac{1}{sind}\left[\frac{sin{\theta }_{2}cos{\theta }_1-cos{\theta }_{2}sin{\theta }_1}{cos{\theta }_{1}cos{\theta }_2}\right]\\ =\frac{1}{sind}\left[\frac{sin{\theta }_{2}cos{\theta }_1}{cos{\theta }_{1}cos{\theta }_2}-\frac{cos{\theta }_{2}sin{\theta }_1}{cos{\theta }_{1}cos{\theta }_2}\right]\\ =\frac{1}{sind}\left[tan{\theta }_2-tan{\theta }_1\right]\\ Similarly,wecansolveothertermswhichwillbe\\ \frac{1}{sind}\left[tan{\theta }_3-tan{\theta }_2\right]and\frac{1}{sind}\left[tan{\theta }_4-tan{\theta }_3\right]\\ HereLHS=\frac{1}{sind}\left[tan{\theta }_2-tan{\theta }_1+tan{\theta }_3-tan{\theta }_2+\dots +tan{\theta }_n-tan{\theta }_{n-1}\right]\\ =\frac{1}{sind}\left[-tan{\theta }_1+tan{\theta }_n\right]=\frac{tan{\theta }_n-tan{\theta }_1}{sind}RHS.\\ LHS=RHSHenceproved.\end{array}$

1. If the sum of  terms of an A.P. is given by , then the common difference of the A.P. is

(a) 3

(b) 2

(c) 6

(d) 4

Sol:

$\begin{array}{l}Giventhat{S}_n=3n+2n^2\\ S_1=3\left(1\right)+2{\left(1\right)}^2=5\\ S_2=3\left(2\right)+2{\left(2\right)}^2=14\\ S_1=a_1=5\\ S_2-S_1=a_2=14-5=9\\ \therefore Commondifferenced=a_2-a_1=9-5=4\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

2. The third term of a G.P. is 4. The product of its first 5 terms is

(a) $4^3$

(b) $4^4$

(c) $4^5$

(d) None of these

Sol:

$\begin{array}{l}Giventhat{T}_3=4\\ \Rightarrow a{r}^{3-1}=4\left[\because T_n=a{r}^{n-1}\right]\\ \Rightarrow a{r}^2=4\\ Productoffirst5terms=a.ar.a{r}^2.a{r}^3.a{r}^4=a^5{r}^{10}={\left(a{r}^2\right)}^5={\left(4\right)}^5\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

1. For $a,b,c$  to be in G.P., the value of $\frac{a-b}{b-c}$ is equal to .............. .

Sol:

$\begin{array}{l}Sincea,bandcareinG.P\\ \therefore \frac{b}{a}=\frac{c}{b}=r\left(constant\right)\\ \Rightarrow b=arandc=br\Rightarrow c=ar.r=a{r}^2\\ So,\frac{a-b}{b-c}=\frac{a-ar}{ar-a{r}^2}=\frac{a\left(1-r\right)}{ar\left(1-r\right)}=\frac{1}{r}=\frac{a}{b}=\frac{b}{c}\\ Hence,thecorrectvalueofthefilleris\frac{a}{b}or\frac{b}{c}.\end{array}$

2. The sum of terms equidistant from the beginning and end in an A.P. is equal to ............ .

Sol:

$\begin{array}{l}LetA.Pbea,a+d,a+2d,a+3d,\dots ,a+\left(n-1\right)d\\ Takingfirstandlastterm\\ a_1+a_n=a+a+\left(n-1\right)d=2a+\left(n-1\right)d\\ Takingsecondandsecondlastterm\\ a_2+a_{n-1}=\left(a+d\right)+\left[a+\left(n-2\right)d\right]=2a+\left(n-1\right)d=a_1+a_n\\ Takingthirdfromthebeginningandthirdfromtheend\\ a_3+a_{n-2}=\left(a+2d\right)+\left[a+\left(n-3\right)d\right]=2a+\left(n-1\right)d=a_1+a_n\\ Fromtheabovepattern,weobservethatthesumoftermsequidistantfromthebeginning\\ andtheendinanA.Pisequaltothe\left[firstterm+lastterm\right]\\ Hence,thecorrectvalueofthefilleris\left[firstterm+lastterm\right].\end{array}$

State whether the statements are True or False:  

1. Two sequences cannot be in both A.P. and G.P. together.

Sol:

$\begin{array}{l}LetusconsideraG.P,a,aranda{r}^2\\ IfitisinA.Pthenar-a\ne a{r}^2-ar\\ Hence,thegivenstatementisTrue.\end{array}$

2. Every progression is a sequence but the converse, i.e., every sequence is also a progression, need not necessarily be true.

Sol:

$\begin{array}{l}Letusconsiderasequenceofprimenumbers2,3,5,7,11,\dots \\ Itisclearthatthisprogressionisasequencebutsequenceisnotaprogression\\ becauseitdoesnotfollowaspecificpattern.\\ Hence,thegivenstatementisTrue.\end{array}$

Sol:

$\begin{array}{l}\left(a\right)4,1,\frac{1}{4},\frac{1}{16}\\ Here,\frac{a_2}{a_1}=\frac{1}{4},\frac{a_3}{a_2}=\frac{1/4}{1}=\frac{1}{4}and\frac{a_4}{a_3}=\frac{1/16}{1/4}=\frac{1}{4}\\ Hence,itisG.P\\ \therefore \left(a\right)\leftrightarrow \left(iii\right)\\ \left(b\right)2,3,5,7\\ Here,a_2-a_1=3-2=1\\ a_3-a_2=5-3=2\\ a_2-a_1\ne a_3-a_2\\ HenceitisnotA.P\\ \frac{a_2}{a_1}=\frac{3}{2},\frac{a_3}{a_2}=\frac{5}{3}\\ So,\frac{3}{2}\ne \frac{5}{3}\\ So,itisnotG.P\\ Hence,itissequence\\ \therefore \left(b\right)\leftrightarrow \left(ii\right)\\ \left(c\right)13,8,3,-2,-7\\ Here,a_2-a_1=8-13=-5\\ a_3-a_2=3-8=-5\\ a_2-a_1=a_3-a_2\\ So,itisanA.P\\ \therefore \left(c\right)\leftrightarrow \left(i\right)\end{array}$

Sol:

$\begin{array}{l}\left(a\right)4,1,\frac{1}{4},\frac{1}{16}\\ Here,\frac{a_2}{a_1}=\frac{1}{4},\frac{a_3}{a_2}=\frac{1/4}{1}=\frac{1}{4}and\frac{a_4}{a_3}=\frac{1/16}{1/4}=\frac{1}{4}\\ Hence,itisG.P\\ \therefore \left(a\right)\leftrightarrow \left(iii\right)\\ \left(b\right)2,3,5,7\\ Here,a_2-a_1=3-2=1\\ a_3-a_2=5-3=2\\ a_2-a_1\ne a_3-a_2\\ HenceitisnotA.P\\ \frac{a_2}{a_1}=\frac{3}{2},\frac{a_3}{a_2}=\frac{5}{3}\\ So,\frac{3}{2}\ne \frac{5}{3}\\ So,itisnotG.P\\ Hence,itissequence\\ \therefore \left(b\right)\leftrightarrow \left(ii\right)\\ \left(c\right)13,8,3,-2,-7\\ Here,a_2-a_1=8-13=-5\\ a_3-a_2=3-8=-5\\ a_2-a_1=a_3-a_2\\ So,itisanA.P\\ \therefore \left(c\right)\leftrightarrow \left(i\right)\end{array}$