Maths

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

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

Then $a_1+a_2+a_3=16$

$a+ar+a{r}^2+16^{}$

$a\left(1+r+r^2\right)=16$ ………. I

Also $a_4+a_5+a_6=128$

$a{r}^3+a{r}^4+a{r}^5=128$

$a{r}^3\left(1+r+r^2\right)=128$ …… II

Dividing II and I we get,

$\frac{a{r}^3\left(1+r+r^2\right)}{a\left(1+r+r^2\right)}=\frac{128}{16}$

r³= 8

r³ = 2³

 r = 2 >1

So, putting r = 2 in I we get,

$a\left(1+2+2^2\right)=16$

$\Rightarrow$ $a\left(1+2+4\right)=16$

$\Rightarrow$ $a\left(7\right)=16$

$\Rightarrow$ $a=\frac{16}{7}$

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

$\Rightarrow$ $\frac{\frac{16}{7}\left(2^n-1\right)}{2-1}$

$\Rightarrow$ $\frac{16}{7}\left(2^n-1\right)$

45. Given, a=3

$r=\frac{3^3}{3}=3$>1

If s_n=120

Then $\frac{a\left(r^n-1\right)}{r-1}=120$

$\Rightarrow$ $\frac{3\left(3^n-1\right)}{3-1}=120$

$\Rightarrow$ $\frac{3\left(3^n-1\right)}{2}=120$

$\Rightarrow$ $3^n-1=120*\frac{2}{3}$

$\Rightarrow$3ⁿ= 80+1

$\Rightarrow$ 3ⁿ= 81

= 3ⁿ = 34

$\therefore r=4$

So, 120 is the sum of first 4 terms of the G.P

44. Let the three terms of the G.P be $\frac{a}{r},a,ar.$

So, $\frac{a}{r}a.ar=1$

= $a^3=1$

= $a=1$

And $\frac{a}{r}+a+ar=\frac{39}{10}$

= $\frac{1}{r}+1+r=\frac{39}{10}$ (as a = 1)

= $\frac{1+r+r^2}{r}=\frac{39}{10}$

= $\left(r^2+r+1\right)*10=39*r$

= $10r^2+10r+10=39r$

= $10r^2+10r-39r+10=0$

= $10r^2-29r+10=0$

= $10r^2-25r-4r+10=0$

= $5r\left(2r-5\right)-2\left(2r-5\right)=10$

$=\left(2x-5\right)\left(5r-2\right)=0$

= $\left(2x-5\right)=0$ or $\left(5r-2\right)=0$

= $r=\frac{5}{2}$ or $r=\frac{2}{5}$

When $a=1,$ $r=\frac{5}{2}$ the terms are,

 $\frac{1}{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.},1,$ $1*\frac{5}{2}=\frac{2}{5},1,\frac{5}{2}$

When $a=1,$ $r=\frac{2}{5}$ the terms are,

$\frac{1}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.},$ 1, $1*\frac{2}{5}=\frac{5}{2},$ 1, $\frac{2}{5}$