Sequence and Series

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, $\frac{2}{5}$

41. Here,  $a_1=1$

$r=\frac{-9}{1}=-a$

So,  $s_n=\frac{a_1\left[1-r^n\right]}{1-r}$

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

= ${\frac{1-\left(a\right)}{1+a}}^n$

39. Here a = 0.15 = $\frac{15}{100}$

$r=\frac{0.015}{0.15}=\frac{\frac{15}{1000}}{\frac{15}{150}}=\frac{1}{10}$ <1

n = 20

So, Sum to term of G.P., $s_n=\frac{a(1-r^n)}{1-r}$

$\Rightarrow$ $s_{20}=\frac{\frac{15}{100}\left[1-{\left(\frac{1}{10}\right)}^{20}\right]}{1-\frac{1}{10}}$

= $\frac{\frac{15}{100}\left[1-{(0.1)}^{20}\right]}{\frac{9}{10}}$

= $\frac{15}{100}*\frac{10}{9}\left[1-{(0.1)}^{20}\right]$

= $\frac{1}{6}\left[1-{(0.1)}^{20}\right]$

38. For $\frac{-2}{7}$ ,  $x$ ,  $\frac{-7}{2}$ to be in G.P. we have the condition,

$\Rightarrow$ $\frac{x}{-2/7}=\frac{-7/2}{x}$

$\Rightarrow$ $x^2=\left(\frac{-7}{2}\right)*\left(\frac{-2}{7}\right)$

$\Rightarrow$ $x^2=1$

$\Rightarrow$ $x=+1$

36. Given, a = 3

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

Then, a₄ = (a₂)²

$\Rightarrow$ar^4-1 =   (ar^2-1)²

$\Rightarrow$ ar³=  (ar)²

$\Rightarrow$ ar³=  a²r²

$\Rightarrow$ $\frac{r^3}{r^2}=\frac{a^2}{2}$

$\Rightarrow$ r = a = 3

$\therefore$ a₇ = ar^7-1  = ar⁶= (-3) (-3)⁶  = (-3)⁷ = -2187.