Ncert Solutions Maths class 11th

99. The given series is.

$\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}$ +…………. upto nth term,

The nth term is ,

$a_n=\frac{1^3+2^3+3^3+........+n^3}{1+3+5+........+n...terms}$ { A.P. of n terms and a = 1, d = 5-3 = 2}

$a_n=\frac{{\left[\frac{n(n+1)}{2}\right]}^2}{\frac{n}{2}[2*1+(n-1)2]}$

$a_n=\frac{\frac{n^2(n+1){}^2}{4}}{\frac{n}{2}[2+(n-1)2]}=\frac{\frac{n^2(n+1){}^2}{4}}{n[1+n-1]}$

$a_n=\frac{\frac{n^2(n+1){}^2}{4}}{n^2}=\frac{{(n+1)}^2}{4}$

$a_n=\frac{n^2+2x+1}{4}=\frac{n^2}{4}+\frac{n}{2}+\frac{1}{4}$

$\therefore Sn={\displaystyle \sum _{}^{}an}=\frac{1}{4}{\displaystyle \sum _{}^{}{x}^2}+\frac{1}{2}{\displaystyle \sum _{}^{}n+{\displaystyle \sum _{}^{}1}}$

$=\frac{1}{4}*\frac{9(n+1)(2n+1)}{6}+\frac{1}{2}*\frac{n(x+1)}{2}+\frac{1}{4}*n$

$=\frac{n}{4}\left[\frac{(n+1)(2n+1)}{6}+(n+1)+1\right]$

$=\frac{n}{4}\left[\frac{2n^2+n(n+1)(2n+1)+6(x+1)+6}{6}\right]$

$=\frac{n}{4}\left[\frac{2n^2+n+2n+1+6n+6+6}{6}\right]$

$=\frac{n}{4}\left[\frac{2n^2+9n+13}{6}\right]$

$=\frac{n\left(2n^2+9n+13\right)}{24}$

98. Given,

S₁ = 1 + 2 + 3 + ………. + n $=n+\frac{(n+1)}{2}$

S₂ = 1² + 2² + 3² + ………… + n²  $=\frac{n(n+1)(2n+1)}{6}$

S₃ = 1³ + 2³ + 3³ + …………. + n³ =${\left[\frac{n(n+1)}{2}\right]}^2$

So, L.H.S. $=9S_2^2=9*{\left[\frac{n(n+1)(2n+1)}{6}\right]}^2$

$=\frac{9*n^2(n+1){}^2(2n+1){}^2}{36}$

$=\frac{n^2(n+1){}^2(2n+1){}^2}{4}$

R.H.S. $=S_3\left(1+8S_1\right)={\left[\frac{n(n+1)}{2}\right]}^2\left[1+8*\frac{n(n+1)}{2}\right]$

$=\frac{n^2(n+1){}^2}{4}*[1+4n(n+1)]$

$=\frac{n^2(n+1){}^2}{4}\left[1+4n^2+4n\right]$

$=\frac{n^2(n+1){}^2}{4}\left[{(2n)}^2+2(2n)1+1^2\right]$

$=\frac{n^2(n+1){}^2}{4}{(2n+1)}^2=$ L.H.S.

So, 9(S₂)² = S₃ ( 1 + 8S₁)

97. The given series is 3+7+13+21+31+ ……. upto n terms

So, Sum, S_n = 3+7+13+21+31+ ………….+ a_n-1 + a_n

Now, taking,

S_n = S_n = [ 3 + 7 + 13 + 21 + 31 + ………. a_n-1 + a_n ] [ 3+ 7 + 21 + 31 + … a_n-1 + a_n]

0 = [ 3+ (7 - 3) + (13 - 7) + (13 - 7) + (21 - 13) + …….+ (a_n a_n-1) - a_n]

0 = 3 + [ 4 + 6 + 8 +…….+ (n-1) terms] a_n

$\Rightarrow a_n=3+\left\{\frac{n-1}{2}[2*4+[(n-1)-1]2]\right\}$ { $?$ 4 + 6 + 8 + ……. is sum of A.P. of n 1 terms with a = 4, d = 6- 4 = 2}

$\Rightarrow a_n=3+\left\{\frac{n-1}{2}[8+(n-2)2]\right\}$

$\Rightarrow a_n=3+\left\{\frac{n-1}{2}*2[4+n-2]\right\}=3+(n-1)(n+2)$

a_n = 3 + n² + (1 + 2) n + (-1) (2) { $?$ (a + b) (a + b) = a2 + (b + c) a + bc }

a_n = 3 + n² + n- 2 = n² + n +1

sum of series, $S_n={\displaystyle \sum _{}^{}{a}_n}={\displaystyle \sum _{}^{}{n}^2}+{\displaystyle \sum _{}^{}n}+{\displaystyle \sum _{}^{}1}$

$=\frac{n(n+1)(2n+1)}{6}+\frac{n(x+1)}{2}+n$

$=n\left[\frac{(n+1){(2n+1)}^{}}{6}+\frac{n+1}{2}+1\right]$

$=n\left[\frac{(n+1)(2n+1)+3(n+1)+6}{6}\right]$

$=n\left[\frac{2n^2+n+2n+1+3n+3+6}{6}\right]$

$=n\left[\frac{2n^2+6n+10]}{6}\right]$

$=n\left[\frac{2\left(n^2+3n+5\right)}{6}\right]$

$=\frac{n\left(n^2+3n+5\right)}{3}$

96. Given, series is $2*4+4*6+6*8+\dots +n$ terms.

So, a₂₀ = ( 20th term of A.P. 2, 4, 6 ….) ( 20th term of A.P. 4,6,8 ………)

i.e. a = 2, d = 4 -2 = 2 i.e. a =4, d = 6- 4 = 2

= [2 + (20- 1)2] [4 + (20-1)2]

= [2+19*2] [4 +19 *2]

= (2+38) (4+38)

= 40*42 = 1680

95. (i) Sn = 5+55+555+…………… upto n term

=5(1+11+111+ …………….upto n term )

Multiplying and dividing by 9 we get

$=\frac{5}{9}(9+99+999+\dots upto\mathrm{nterms})$

 

$=\frac{5}{9}\left[\right(10+10^2+10^3+?$ upto\mathrm{nterms}$$
$-n*1]$

$=\frac{5}{9}\left[\frac{10\left(10^n-1\right)}{10-1}-n\right]$ { $?10+10^2+10^3+\dots \mathrm{ntermsisaG.Pof}a=10,r=10>1\mathrm{nterms}$ }

$=\frac{5}{9}*\left[\frac{10}{9}\left(10^n-1\right)-n\right]$

$=\frac{50}{81}\left(10^n-1\right)-\frac{5n}{9}$

(ii) $S_n=0.6+0.66+0.666+\dots n$

$=6[0.1+0.11+0.111+?uptonterms$$]$

$=\frac{6}{9}[0.9+0\cdot 99+0.999+?\mathrm{<math><mrow><mo>uptonterms</mo></mrow></math>}]$ {multiplying and dividing by 9}

$=\frac{6}{9}[(1+1+1\dots n)-(0.1+0.01+0.001+\dots \mathrm{<math><mrow><mo>uptonterms</mo></mrow></math>})]$

$=\frac{6}{9}\left[n-\left(\frac{1}{10}+\frac{1}{10}*\frac{1}{10}+\frac{1}{10}*{\left(\frac{1}{10}\right)}^2+?\mathrm{<math><mrow><mo>uptonterms</mo></mrow></math>}\right)\right]$

$=\frac{6}{9}\left[n-\frac{\frac{1}{10}\left(1-{\left(\frac{1}{10}\right)}^n\right)}{1-\frac{1}{10}}\right]$

$=\frac{6}{9}\left[n-\frac{1-{\left(\frac{1}{10}\right)}^n}{10-1}\right]$

$=\frac{6}{9}\left[x-\frac{1}{9}\left(1-\frac{1}{10^n}\right)\right]$

$=\frac{6}{9}*\frac{1}{9}\left[9x-\left(1-\frac{1}{10^n}\right)\right]$

$=\frac{2}{27}\left[9n-\left(1-\frac{1}{10^n}\right)\right]$

$=\frac{2}{27}\left[9n-1+\frac{1}{10^n}\right]$

94. As a, b, c are in A.P. we can write,

$b-a=c-b$

$\Rightarrow b+b=a+c$

$\Rightarrow 2b=a+c$ I

As b, c, d are in G.P. we can write,

$\frac{c}{b}=\frac{d}{c}$

$\Rightarrow c^2=bd$ II

And ar $\frac{1}{c},\frac{1}{d},\frac{1}{e}$ are in A.P. we can write,

$\Rightarrow \frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}$

$\Rightarrow \frac{1}{d}+\frac{1}{d}=\frac{1}{c}+\frac{1}{e}$

$\Rightarrow \frac{2}{d}=\frac{e+c}{ce}$

$\Rightarrow \frac{d}{2}=\frac{ce}{e+c}$

$\Rightarrow d=\frac{2ce}{c+e}$ …………. III

Now, $c^2=Bd$ from II

$\Rightarrow =\frac{a+c}{2}*\frac{2ce}{c+c}$ {from 1 and 3}

$\Rightarrow c^2=\frac{(a+c)ce}{c+e}$

$\Rightarrow c=\frac{(a+c)e}{(c+e)}$

$\Rightarrow c(c+e)=(a+c)e$

$\Rightarrow c^2+ce=ac+ce$

$\Rightarrow c^2=ae$

$\Rightarrow \frac{c}{a}=\frac{e}{c}$

i.e., a, c and e are in G.P.

92. Given, ab are roots of $x^2-3x+p=0$

and c & d are roots of $x^2-12x+q=0$

So, $\Rightarrow a+b=\frac{-\left(-3\right)}{1}$ and $ab=\frac{P}{1}$

a + b = +3 ………….I and ab = P…………….II { $?$ sum of roots = $\frac{-B}{A}$ , Product of roots = $\frac{-C}{A}$ }

Similarly, c + d $=\frac{-(-12)}{1}$ and $cd=\frac{q}{1}$

c + d = 12 ……….III ad cd = q (4)

As a, b, c, d from a G.P and if r be the common ratio

a = a

b = ar

c = ar²

d = ar³

So, from equation, (1),

$a+b=3\Rightarrow a+ar=3\Rightarrow a(1+r)=3$ (5)

And $c+d=12\Rightarrow a{r}^2+a{r}^3=12\Rightarrow a{r}^2(1+r)=12$ (6)

Dividing equation (6) and (5) we get,

$\frac{a{r}^2(1+r)}{a(1+r)}=\frac{12}{3}$

 r² = 4

Now, L.H.S. $=\frac{q+p}{q-p}=\frac{cd+ab}{cd-ab}$ {from (4) and (5)}

$=\frac{a{r}^2*a{r}^3+a*ar}{a{r}^2+a{r}^3-a*ar}$

$=\frac{a^{2}r\left(r^4+1\right)}{a^{2}r\left(r^4-1\right)}$

$=\frac{{\left(r^2\right)}^2+1}{{\left(r^2\right)}^2-1}=\frac{4^2+1}{4^2-1}=\frac{16+1}{16-1}=\frac{17}{15}$ = R.H.S.

91. Let r be the common ratio of the G.P.

Then, a, b, c, d a, ar, ar², ar³

So, $a^n+b^n=a^n+{(ar)}^n=a^n+a^n{r}^n=a^n\left(1+r^n\right)-(1)$

$b^n+c^n={(ar)}^n+{\left(a{r}^2\right)}^n=a^n{r}^n+a^n{r}^{2n}=a^n{r}^n\left(1+r^n\right)$ (2)

$c^n+d^n={\left(a{r}^2\right)}^n+{\left(a{r}^3\right)}^n=a^n{r}^{2n}+a^n{r}^{3n}=a^n{r}^{2n}\left(1+r^n\right)$ (3)

Hence, $\frac{b^n+c^n}{a^n+b^n}=\frac{a^n{r}^n\left(1+r^n\right)}{a^n\left(1+r^n\right)}=r^n$ {from (2) and (1)}

and $$ $\frac{c^n+d^n}{b^n+c^n}=\frac{a^n{r}^{2n}\left(1+r^2\right)}{a^n{r}^n\left(1+r^2\right)}=r^n$ {from (3) and (2)}

i.e., $\frac{b^n+c^n}{a^x+b^n}=\frac{c^n+d^n}{b^x+c^n}$

$\therefore a^n+b^n,b^n+c^n,c^n+d^n$ are in A.P

90. Given, $a\left(\frac{1}{b}+\frac{1}{c}\right),b\left(\frac{1}{c}+\frac{1}{a}\right),c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in A.P.

So, $\frac{a(c+b)}{bc},\frac{b(a+c)}{ac},\frac{c(b+a)}{ab}$ are in A.P.

$\Rightarrow \frac{ac+ab}{bc},\frac{ab+bc}{ac},\frac{bc+ac}{ab}$ are in A.P.