Class 11th

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.

If we add 1 to all each terms of the sequence it will given be an A.P of common difference 1.

So, $\frac{ac+ab}{bc}+1,\frac{ab+bc}{ac}+1,\frac{bc+ac}{ab}+1$ are in A.P.

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

Dividing add of the sum by ab + bc + ac will conserve.

then A.P so,

$\Rightarrow \frac{1}{bc},\frac{1}{ac},\frac{1}{ab}$ are in A.P.

Similarly multiplying each term by abc we get,

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

a, b, c, are in A.P.

Hence proved

89. Let A and d be the first term & common difference of the A.P.

Then,

$ap=a\to A+(p-1)d=a$ ………I

$a_r=c\to A+(M-1)d=c$ …………III

So, L.H.S. $=(q-1)a+(n-p)b+(q-q)c$

$=(q-r)[A+(p-1)d]+(n-p)[A+(q-1)d]+(p-q)[A+(r-1)d]$

{putting value for I, II, III}

$\Rightarrow Aq+q(p-1)d-Ar-r(p-1)d+Ar+r\left(q-1\right)d-Ap-p\left(q-1\right)d$

$+Ap+p\left(r-1\right)d-Aq-q\left(r-1\right)d$

$\Rightarrow pqd-qd-rpd+rd+rqd-rd-pqd+pd+rpd-pd-rqd+qd$

$=0=$ R.H.S

88. Let a and r be the first term & common ratio of the G.P.

So, S = a +ar + ar² +……… upto n terms.

$S=\frac{a\left(1-r^n\right)}{1-r}$

and P = a .ar. ar² ar . upton n terms.

$=a^n{r}^{1+2+3+\dots +(n-1)}$

$=a^{n}r\frac{(n-1)(n-1+1)}{2}$

$=a^{n}r\frac{n(n-1)}{2}$

And R = sum of reciprocal of n terms ( $\frac{1}{a}+\frac{1}{a{r}^n}+...........$ upto n terms)

$=\frac{\frac{1}{a}\left[{\left(\frac{1}{r}\right)}^n-1\right]}{\frac{1}{r}-1}$  As r <1

$\frac{1}{r}$ >1

$=\frac{\frac{1}{a}\left[\frac{1}{r^n}-1\right]}{\frac{1-r}{r}}=\frac{1}{a}\left[\frac{1-r^n}{r^n}\right]*\frac{r}{1-r}$

$=\frac{1-r^n}{a{r}^n}*\frac{r}{1-r}$

$=\frac{1-r^n}{a(1-r)r^{n-1}}$ …. III

Now, L.H.S. = P² Rⁿ

$={\left[a^{n}r\frac{n\left(n-1\right)}{2}\right]}^2·{\left[\frac{1-x^n}{a(1-n)r^{n-1}}\right]}^n$ { equation II & III}

$=a^{2n}\cdot r^{n(n-1)}*\frac{{\left[1-r^n\right]}^n}{a^n(1-n){}^n}{r}^{n(n-1)}$

$=a^{2x-n}*\frac{r^{n(x-1)}}{r^{n(n-1)}}*\frac{{\left[1-x^n\right]}^n}{{(1-r)}^n}$

$=\frac{a^n{\left[1-r^n\right]}^n}{{(1-r)}^n}$

$={\left[\frac{a\left[1-r^n\right]}{(1-r)}\right]}^n$

$=S^n=$R.H.S { $?$ equation I}

87. Here, $\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}$

$\Rightarrow (a+bx)(b-cx)=(b+cx)(a-bx)$

$\Rightarrow ab-acx+b^{2}x-bc{x}^2=ab-b^{2}x+acx-bc{x}^2$

$\Rightarrow ab-ab-acx-acx=-b^{2}x-b^{2}x+bc{x}^2-bc{x}^2$

$\Rightarrow -2acx=-2b^{2}x$

$\Rightarrow ac=b^2$

$\Rightarrow \frac{c}{b}=\frac{b}{a}$ …………I

And $\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}$

$\Rightarrow (b+cx)(c-dx)=(c+dx)(b-cx)$

$\Rightarrow bc-bdx+c^{2}x-cdx=bc-c^{2}x+bdx-cdx$

$\Rightarrow bc-bc+c^{2}x+c^{2}x=bdx+bdx-cdx+cdx$ $$

$\Rightarrow 2c^{2}x=2bdx$

$\Rightarrow c^2=bd$

$\Rightarrow \frac{c}{b}=\frac{d}{c}$ ……………II

From I and II

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

a, b, c and d are in G.P.