Class 11th

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.

86. Given, a = 11

Let d and l be the common difference & last term of the A.P.

Then, $a+(a+d)+(a+2d)+(a+3d)=56$ [first 4 terms sum]

$\Rightarrow 4a+6d=56$

$\Rightarrow 6d=56-4a=56-4*11=56-44=12$

$\Rightarrow d=\frac{12}{6}=2$

And, $l+(l-d)+(l-2d)+(l-3d)=112$

$\Rightarrow 4l-6d=112$

$\Rightarrow 4l=112+6d=112+6*2=112+12=124$ [last 4 terms sum]

$\Rightarrow l=\frac{124}{4}=31$

So, $l=31$

$\Rightarrow a+{(n-1)}^d=31$

$\Rightarrow 11+{(n-1)}^2=31$

$\Rightarrow (n-1)2=31-11=20$

$\Rightarrow n-1=\frac{20}{2}=10$

$\Rightarrow n=10+1$

$\Rightarrow n=11$

the A.P. has 11 number of terms.

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

Then, number of term = 2n (even).

$a_1+a_2+?+a_{2n}=5\left(a_1+a_3+?+a_{2n-1}\right)[?]$

$\Rightarrow a+ar+.........+a{r}^{2n-1}=5\left(a+a{r}^2+.......+a{r}^{2n-1-1}\right)$

$\Rightarrow \frac{a\left(1-r^{2n}\right)}{1-n}=5*\frac{a\left[1-{\left(r^2\right)}^n\right]}{1-r^2}$ { series on R.H.S. has $\frac{2n}{2}=n$ terms and common ratio $\frac{a{r}^2}{a}=r^2$ }

$\Rightarrow \frac{1-r^{2n}}{1-r}=\frac{5\left[1-r^{2n}\right]}{1-r^2}$ (eliminating a)

$\Rightarrow \frac{1-r^{2r}}{1-r}=\left[\frac{1-r^{2r}}{1-r}\right]*\frac{5}{1+r}\left\{?a^2-b^2=(a-b)(a+b)\right\}$ 

$1=\frac{5}{1+r}\{Eliminating\; same\; term\}$

$\Rightarrow 1+r=5$

$\Rightarrow r=5-1$

r = 4

84. Let a, ar and $a{r}^2$ be the three nos. which is in G.P.

Then, a + ar + ar² = 56

a ( 1 + r + r²) =56  -I

Given, that a1, ar 7, ar² - 21 from an AP we have,

$(ar-7)-(a-1)=\left(a{r}^2-21\right)-(ar-7)$

$\Rightarrow ar-7-a+1=a{r}^2-21-ar+7$

$\Rightarrow ar-a-6=a{r}^2-ar-14$

$\Rightarrow a{r}^2-ar-ar+a-a-14-6$

$\Rightarrow a{r}^2-2ar+a=8$

$\Rightarrow a\left(r^2-2r+1\right)=8$ ………………. II

Now, dividing equation I by II we get,

$\Rightarrow \frac{a\left(1+n+n^2\right)}{a\left(r^2-2r+1\right)}=\frac{56}{8}$

$\Rightarrow 1+r+r^2=7\left(n^2-2n+1\right)$

$\Rightarrow 1+r+r^2=7r^2-14n+7$

$\Rightarrow 7r^2-14n+7-1-x-r^2=0$

$\Rightarrow 6r^2-15r+6=0$

$\Rightarrow 2r^2-5r+2=0$ (dividing by 3 throughout)

$\Rightarrow 2r^2-4r-r+2=0$

$\Rightarrow 2r(r-2)-(r-2)=0$

$\Rightarrow (r-2)(2r-1)=0$

$\Rightarrow r-2=02r-1=0$

$\Rightarrow r=2r=\frac{1}{2}$

So, when r = 2, putting in equation I,

$\Rightarrow a\left(1+2+2^2\right)=56$

$\Rightarrow a(1+2+4)=56$

$\Rightarrow a(7)=56$

$\Rightarrow a=\frac{56}{7}=8$

The numbers are 8, 8* 2, 8* 2² = 8, 16, 32.

And When $r=\frac{1}{2}$ putting in equation I,

$a\left(1+\frac{1}{2}+\frac{1}{2^2}\right)=56$

$\Rightarrow a\left(1+\frac{1}{2}+\frac{1}{4}\right)=56$

$\Rightarrow a\left(\frac{4+2+1}{4}\right)=56$

$\Rightarrow a*\frac{7}{4}=56$

$\Rightarrow a=56*\frac{4}{7}=32$

So, the numbers are $32,32*\frac{1}{2},32*{\left(\frac{1}{2}\right)}^2\Rightarrow 32,16,8$