Class 11th

55. Given, first term and xth term and a and b

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

Then,

Product of x terms, p= $\left(a\right)$ $\left(ar\right)$ $\left(a{r}^2\right)$ …… $\left(a{r}^{n-1}\right)$

p= $\left(a\right)$ $\left(ar\right)\left(a{r}^2\right)$ …. $\left(a{r}^{n-2}\right)\left(a{r}^{n-1}\right)$

= (a *a* ….n term)(r´r²´r³ …. $r^{n-2}*r^{n-1}$ )

= $a^x$ $r^{\left[1+2+3+....\left(n-2\right)+\left(n-1\right)\right]}$

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

So, $p^2={\left[a^n{r}^n\left(\frac{n-1}{2}\right)\right]}^2$ [We know that,

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

= $a^{2n}{r}^{2n}\left(\frac{n-1}{2}\right)$ [So, $1+2+3+....+\left(n-1\right)=\frac{\left(n-1+1\right)\left(n-1\right)}{2}=\frac{n(n-1)}{2}$

= $a^{2n}*r^{n\left(n-1\right)}$

= ${\left(a*a{r}^{n-1}\right)}^n$ $[?$ $a{n}^{n-1}=$ last term = b (given)]

= ${\left(a*b\right)}^n$

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

Then, ap = a

$A{R}^{p-1}=a$ …… I

and aq = b

$A{R}^{q-1}=b$ …….II

Also, ar = c

$A{R}^{r-1}=c$ ….III

Given, L.H.S. = $a^{a-r}$ $b^{r-p}$ $c^{p-q}$

${\left(A{R}^{p-1}\right)}^{q-r}$ ${\left(A{R}^{q-1}\right)}^{r-p}$ $\left(A{R}^{r-1}\right)$ $?$ using I, II and III

$A^{q-r}$ $R^{\left(p-1\right)}$ $A^{r-p}$ $R^{\left(q-1\right)\left(r-p\right)}$ $A^{p-q}$ $R^{\left(r-1\right)\left(p-q\right)}$

$A^{\left[q-r+r-p+p-q\right]}$ $R^{\left[\left(p-1\right)\left(q-r\right)+\left(q-1\right)\left(n-p\right)+\left(x-1\right)\left(p-q\right)\right]}$

A0 R[pqprq+r+qrpqr+p+prqrp+q] $$

$1*$ $R^{\left[pq-pq+pr-pr+q-q+r-r+qr-qr+p-p\right]}$

= R

= 1

= R.H.S

53. Let the four numbers is G.P. be a, ar, ar², ar³

Given, $a{r}^2-a=9$

$\Rightarrow$ $a\left(r^2-1\right)=9$ I

And $\Rightarrow$ $ar-a{r}^3=18$

$\Rightarrow$ $ar\left(1-r^2\right)=18$

$\Rightarrow$ $\left(-1\right)ar\left(x^2-1\right)=18$

$\Rightarrow$ $ar\left(x^2-1\right)=-18$ II

Dividing II by I we get,

$\frac{ar\left(r^2-1\right)}{a\left(r^2-1\right)}=\frac{-18}{9}$

r = 2

So, putting r = 2 in I we get ,

$\Rightarrow$ $a\left[{\left(-2\right)}^2-1\right]=9$

$\Rightarrow$ $a\left[4-1\right]=9$

$\Rightarrow$ $a*3=9$

$\Rightarrow$ $a\frac{9}{3}$

$\Rightarrow$ a = 3

$\therefore$ The four numbers are 3, 3´(-2), 3´(-2)², 3´(-2)³

$\Rightarrow$ 3, 6, 12, 24

The product of corresponding terms of the given sequence are

= $a*A+\left(ar\right)*AR+\left(a{r}^2\right)*A{R}^2+......\left(a{r}^{x?1}\right)\left(A{R}^{x?1}\right)$

= $aA+\left(aA\right)+\left(rR\right)+\left(aA\right)$ $......+\left(aA\right){\left(rR\right)}^{x?1}$

So, looking at the sequence forms a G.P.

(above the)

With common ratio = $\frac{\left(aA\right)\left(rR\right)}{\left(aA\right)}=rR$

51. The sum of product of corresponding terms of the given

Sequences = $\left(2*128\right)+\left(4*32\right)+\left(8*8\right)+\left(16*2\right)+\left(32*\frac{1}{2}\right)$

= 256+128+69+32+16

The above is a G.P. of a = 256,  $r=\frac{128}{256}=\frac{1}{2}$< 1 and x = 5

Sum required = $\frac{256\left(1-{\left(\frac{1}{2}\right)}^5\right)}{1-\frac{1}{2}}$

= $\frac{256\left(1-\frac{1}{32}\right)}{\frac{2-1}{2}}$

= $\frac{256*\left(\frac{32-1}{32}\right)}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=256*2*\frac{31}{32}=496$

50. The given sequence, 8, 88, 888, 8888, …., upto xterm is not a G.P. so we can such that it will be changed to a G.P. by the following .

Sum of x terms, s_x = 8+ 88+888+8888 ….upto x term

= $8\left[1+11+111+1111+......\right]$ upto x term

Multiplying the numerator and denumerator by 9 we get,

= $\frac{8}{9}\left[9+99+999+9999+.........\right]$ x terms

= $\frac{8}{9}\left[\left(10-1\right)+\left(10^2-1\right)+\left(10^3-1\right)+\left(10^4-1\right)+.....\right]$ upto x terms

= $\frac{8}{9}$ [(10 + 1010 + 103 +104 …….upto, x terms) (1 +1 + 1 + 1 …… upto, x terms)]

= $\frac{8}{9}\left[\frac{10\left(10^n-1\right)}{10-1}-1*n\right]$

= $\frac{8}{9}\left[\frac{10\left(10^n-1\right)}{9}-n\right]$

= $\frac{80}{81}\left(10^n-1\right)-\frac{8}{9}n$

49. Let aand r be the first term and common ration of the G.P.

So,  $a_4=x$

$\Rightarrow$ $a{r}^3=x$ - I

And a₁₀ = y

$\Rightarrow$ $a{r}^9=y$ - II

Also a 16 = z

$a{r}^{15}=z$ - III

So,  $\frac{y}{x}=\frac{a{r}^9}{a{r}^3}=r^{9-3}=r^6$

and $\frac{z}{y}=\frac{a{r}^{15}}{a{r}^9}=r^{15-9}=r^6$

$?\frac{y}{x}=\frac{z}{y}$

$?$ x, y, z are in G.P

48. Let a and r be the first term and common ratio.

Then, $\Rightarrow$ $a_1+a_2=-4$

$\Rightarrow$ $a+ar=-4$

$\Rightarrow$ $a\left(1+r\right)=-4$

$\Rightarrow$ a (1+ n) = 4

$\Rightarrow$ $a{r}^4=4a{r}^2$

$\Rightarrow$ r² = 4

$\Rightarrow$ r² =2²

$\Rightarrow$ r = ±2

When r =2

a (1+2)=-4

a 3 = -4

$a=\frac{-4}{3}$

So, the G.P. is a, ar, ar²,…………. $\Rightarrow$ $\frac{-4}{3},\frac{-4}{3}*2,\frac{-4}{3}{\left(2\right)}^2...........$

$\Rightarrow$ $\frac{-4}{3},\frac{-8}{3},\frac{-16}{3},.........$

When $r=-2$

$a=\left(1-2\right)=-4$

a = (-) = -4

a = 4

So, the reqd. by G.P. is a, ar, ar²…………. $\Rightarrow$ 4, 4 (-2), 4 (-2)², .

$\Rightarrow$ 4, 8, 16, .

47. Given, a= 729

$a_7=a{r}^6=64$

$\Rightarrow$ 729. r⁶ = 64

$\Rightarrow$ $r^6=\frac{64}{729}$

$\Rightarrow$ $r^6={\left(\frac{2}{3}\right)}^6$

$\Rightarrow$ $r=+\frac{2}{3}$ <1

When $r=\frac{2}{3}$

= $s_7=\frac{a\left(1-r^7\right)}{4-r}=\frac{729\left(1-{\left(\frac{2}{3}\right)}^7\right)}{1-\frac{2}{3}}=\frac{729\left[1-\frac{128}{2187}\right]}{\frac{3-2}{2}}$

= $729*\frac{3}{1}*\left[\frac{2187-128}{2187}\right]$

= $729*\frac{3*2059}{2187}$

= 2059

When $r=\frac{2}{3}$

$s_7=\frac{a\left(1-r^7\right)}{1-r}=\frac{729\left[1-{\left(\frac{-2}{3}\right)}^7\right]}{1-\left(\frac{-2}{3}\right)}=\frac{729\left[1+\frac{128}{2187}\right]}{1+\frac{2}{3}}$

= $\frac{729\left[\frac{2187+128}{2187}\right]}{\frac{3+2}{3}}$

= $729*\frac{3}{5}*\left[\frac{2315}{2187}\right]$

= 463

(i) Due to inert pair effect Pb²⁺ is more stable than Pb⁴⁺. Whereas Sn⁴⁺ is more stable than Sn²⁺.
(ii) 3Ga⁺ —–> 2Ga + Ga³⁺
This is because Ga³⁺ is more stable than Ga⁺