Infinite Sum of Geometric Series : Part (1)
Geometric Progression တစ်ခု၏ $n$ terms အထိ ပေါင်းလဒ်ကို
$S_n= \dfrac{a(1-r^n)}{1-r}$
ဟုသိရှိခဲ့ပြီး ဖြစ်သည်။ ညီမျှခြင်းကို အောက်ပါအတိုင်း အကျယ်ဖြန့်ကြည့်ပါမည်။
$\begin{aligned}
S_n&= \dfrac{a(1-r^n)}{1-r}\\\\
&= \dfrac{a-ar^n}{1-r}\\\\
&= \dfrac{a}{1-r}-\dfrac{ar^n}{1-r}\\\\
\end{aligned}$
Case I.
$|r|>1 \text{ i.e., } r<-1 \text{ or } r>1,$
$(\text{ e.g., } a=2, r=2)$
$\begin{aligned} S_n &= \dfrac{2}{1-2}-\dfrac{2\times2^n}{1-2}\\\\ &= -2+2\times2^n\\\\ &= 2(2^n - 1)\\\\ \end{aligned}$ $\begin{array}{|r|r|} \hline n\hspace{.2cm} & S_n\hspace{4cm} \\ \hline 1 & 2 \\ \hline 5 & 62 \\ \hline 10 & 2046 \\ \hline 50 & 2251799813685246 \\ \hline 100 & 2535301200456458802993406410750 \\ \hline \end{array}$ |
|---|
ထိုအခါ $n$ ၏ တန်ဖိုးကြီးလာသည်နှင့် အမျှ $S_n$ ၏ ပမာဏ (magnitude) သည်လည်း ကြီးလာပါမည်။
$n\rightarrow \infty$ ($n$ သည် အနန္တပမာဏသို့ ချဉ်းကပ်သွားသည်)
$S_n\rightarrow \infty$ ($S_n$ သည်လည်း အနန္တပမာဏသို့ ချဉ်းကပ်သွားသည်)
ထို့ကြောင့် အနန္တကိန်းလုံးအရေအတွက်ထိ ပေါင်းလဒ်ကို ရှာယူရန် မဖြစ်နိုင်တော့ပါ။
ထိုအခြေအနေကို divergent condition ဟုခေါ်သည်။ Divergent Condition တွင် Infinite Sum (အနန္တကိန်းလုံးအရေအတွက်ထိ ပေါင်းလဒ်)
ကို ရှာ၍မရနိုင်ပါ။
Case II.
$|r|=1 \text{ i.e., } r=\pm 1$
အကယ်၍ $r=1$ ဖြစ်လျှင်
$S_n = a + a + a + \dots$ to $n$ terms = $na$
ထိုအခါ $n$ ၏ တန်ဖိုးကြီးလာသည်နှင့် အမျှ $S_n$ ၏ ပမာဏ (magnitude) သည်လည်း ကြီးလာပါဦးမည်။
ထို့ကြောင့် အနန္တကိန်းလုံးအရေအတွက်ထိ ပေါင်းလဒ်ကို ရှာယူရန် မဖြစ်နိုင်တော့ပါ။
အကယ်၍ $r=-1$ ဖြစ်လျှင်
$S_n = a - a + a -a + a - \dots$ to $n$ terms = $a \text { if } n \text { is odd}$
$S_n = a - a + a -a + a - \dots$ to $n$ terms = $0 \text { if } n \text { is even}$
ထို့ကြောင့် အနန္တကိန်းလုံးအရေအတွက်ထိ ပေါင်းလဒ်ကို ရှာယူရန် မလိုတော့ပါ။
Case III.
$|r|<1 \text{ i.e., } -1< r < 1 $
$ (\text{ e.g., } a=2, r=\dfrac{1}{2})$
$\begin{aligned} S_n &= \dfrac{2}{1-1/2}-\dfrac{2\times(1/2)^n}{1-1/2}\\\\ &= 4-4{\left( {\dfrac{1}{2}} \right)}^{n}\\\\ &= 4\left[ {1-{{{\left( {\dfrac{1}{2}} \right)}}^{n}}} \right]\\\\ \end{aligned}$ $\begin{array}{|r|l|r|} \hline n\hspace{.2cm} & \hspace{1.5cm}r^n & S_n\hspace{2.5cm} \\ \hline 1 & 0.5 & 2 \\ \hline 2 & 0.25 & 3 \\ \hline 3 & 0.125 & 3.5 \\ \hline 4 & 0.0525 & 3.75 \\ \hline 5 & 0.03125 & 3.875 \\ \hline 10 & 0.0009765625 & 3.9960937500000000000 \\ \hline 50 & 8.88\times 10^{-16}\approx 0 & 3.9999999999999964473 \\ \hline 100 & 7.88\times 10^{-31}\approx 0 & 4.0000000000000000000 \\ \hline 1000 & 9.33\times 10^{-32}\approx 0 & 4.0000000000000000000 \\ \hline \end{array}$ |
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ထိုအခါ $n$ ၏ တန်ဖိုးကြီးလာသည်နှင့် အမျှ $S_n$ ၏ ပမာဏ (magnitude) သည် တန်ဖိုးတစ်ခုတွင် မပြောင်းလဲတော့ဘဲ ကိန်းသေ ဖြစ်လာပါသည်။
$|r|< 1$ အတွက် $n\rightarrow \infty$ ဖြစ်သည့်အခါ $S_n$ သည် တန်ဖိုးတစ်ခုတွင် မပြောင်းမလဲ တည်ရှိနေသည်။ အဆိုပါတန်ဖိုးမှာ
limit of $S_n$ as $n$ approaches $\infty$ ပင် ဖြစ်သည်။
$\begin{aligned}
\lim _{n \rightarrow \infty} S_{n} &=\lim _{n \rightarrow \infty}\left(\dfrac{a}{1-r}-\dfrac{a r^{n}}{1-r}\right) \\
&=\dfrac{a}{1-r} \times \lim _{n \rightarrow \infty}\left(1-r^{n}\right) \\
&=\dfrac{a}{1-r}(1-0) \quad(\because\ |r|<1)\\
&=\dfrac{a}{1-r}
\end{aligned}$
အထက်ဖေါ်ပြပါ geometric series အမျိုးအစားကို convergent series ဟုခေါ်ပြီး Converget Geometric Series တစ်ခုတွင် infinite
sum (Sum to infinity) ရှိသည်ဟု မှတ်ယူရမည်။ Sum to Infinity ကို သင်္ကေတအားဖြင့် $(S)$ ဟုသတ်မှတ်သည်။
Exercises
- Determine whether the sum to infinity for each of the following geometric
progression exist and find the sum to infinity where they exist.
$\begin{aligned}
&\\
\text{(a)} &\ 3,0.3,0.03, \ldots \\\\
\text{(b)} &\ \dfrac{1}{2}, \dfrac{2}{3}, \dfrac{8}{7}, \dfrac{32}{27}, \ldots \\\\
\text{(c)} &\ \dfrac{7}{2}, 3, \dfrac{18}{7}, \dfrac{108}{49}, \ldots\\\\
\end{aligned}$
SOLUTION
(a) $3,0.3,0.03, \ldots$ is a G.P.
$\begin{aligned}
&\\
\therefore\ a &=3 \\\\
r &=\dfrac{0.3}{3}=\dfrac{1}{10} \\\\
|r| &=\dfrac{1}{10}< 1\\\\
\end{aligned}$
$\therefore$ Sum to infinity exists.
$\begin{aligned}
&\\
S &=\dfrac{a}{1-r} \\\\
&=\dfrac{3}{1-\dfrac{1}{10}} \\\\
&=\dfrac{10}{3}\\\\
\end{aligned}$
(b) $\dfrac{1}{2}, \dfrac{2}{3}, \dfrac{8}{9}, \dfrac{32}{27}, \ldots$ is a G.P.
$\begin{array}{l}
\\
a=\dfrac{1}{2} \\\\
r=\dfrac{2 / 3}{1 / 2}=\dfrac{4}{3} \\\\
\therefore\ |r|=\dfrac{4}{3}>1\\\\
\end{array}$
$\therefore$ Sum to infinity doesn't exist.
$\dfrac{7}{2}, 3, \dfrac{18}{7}, \dfrac{108}{49}, \ldots$ is a G.P.
$\begin{array}{l}
\\
a=\dfrac{7}{2} \\\\
r=\dfrac{3}{(7 / 2)}=\dfrac{6}{7} \\\\
|r|=\dfrac{6}{7}<1\\\
\end{array}$
$\therefore$ Sum to infinity exists.
$\begin{aligned}
&\\
S &=\dfrac{a}{1-r} \\\\
&=\dfrac{7 / 2}{1-\dfrac{6}{7}} \\\\
&=\dfrac{7}{2} \times 7 \\\\
&=\dfrac{49}{2}
\end{aligned}$
- Find the sum to infinity of the following geometric series.
$\begin{aligned}
&\\
\text{(a)} &\ 2+\dfrac{4}{3}+\dfrac{8}{9}+\ldots\\\\
\text{(b)} &\ 2+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32} \ldots\\\\
\text{(c)} &\ 3-\dfrac{2}{3}+\dfrac{4}{27}+\ldots\\\\
\text{(d)} &\ 81-27+9-3+\ldots\\\\
\end{aligned}$
SOLUTION
\begin{aligned}
\text{(a) }\quad 2 &+\dfrac{4}{3}+\dfrac{8}{9}+\cdots \\\\
a &=2 \\\\
r &=\dfrac{4 / 3}{2}=\dfrac{2}{3} \\\\
S &=\dfrac{a}{1-r} \\\\
&=\dfrac{2}{1-\dfrac{2}{3}} \\\\
&=\dfrac{2}{1 / 3} \\\\
&=6
\end{aligned}
$\begin{aligned}
\text { (b) }\quad 2 &+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+\cdots \\\\
a &=2 \\\\
r &=\dfrac{1 / 2}{2}=\dfrac{1}{4} \\\\
S &=\dfrac{a}{1-r} \\\\
&=\dfrac{2}{1-\dfrac{1}{4}} \\\\
&=2 \times \dfrac{4}{3} \\\\
&=\dfrac{8}{3}
\end{aligned}$
$\begin{aligned}
&\\\\
\text { (c) }\quad &3 -\dfrac{2}{3}+\dfrac{4}{27}-\cdots \\\\
a &=3 \\\\
r &=\dfrac{-2 / 3}{3}=-\dfrac{2}{9} \\\\
S &=\dfrac{a}{1-r} \\\\
&=\dfrac{3}{1+\dfrac{2}{9}} \\\\
&=3 \times \dfrac{9}{11} \\\\
&=\dfrac{27}{11}
\end{aligned}$
$\begin{aligned}
&\\\\
\text { (d) }\quad &81 -27+9-3+\cdots \\\\
a &=81 \\\\
r &=-\dfrac{27}{81}=-\dfrac{1}{3} \\\\
S &=\dfrac{a}{1-r} \\\\
&=\dfrac{81}{1+\dfrac{1}{3}} \\\\
&=81 \times \dfrac{3}{4} \\\\
&=\dfrac{243}{4}
\end{aligned}$
- The $3^{\text {rd }}$ and $6^{\text {th }}$ term of a geometric progression are $9$ and
$2 \dfrac{2}{3}$ respectively. Calculate the common ratio, the first term and the sum to
infinity of the progression.
SOLUTION
$\begin{aligned}
\text { In a G.P., }&\\\\
u_{3}&=9 \\\\
a r^{2}&=9 \ldots(1)\\\\
u_{6}&=2 \dfrac{2}{3}\\\\
a r^{5}&=\dfrac{8}{3} \cdots(2) \\\\
\text{B y } (2) \div(1)& \\\\
\dfrac{a r^{5}}{a r^{2}}&=\dfrac{8 / 3}{9}\\\\
r^{3}&=\dfrac{8}{27} \\\\
r&=\dfrac{2}{3}\\\\
\text { Substitute } r&=\dfrac{2}{3} \text { in equation }(1), \\\\
a\left(\dfrac{2}{3}\right)^{2}&=9 \\\\
a&=\dfrac{81}{4} \\\\
S&=\dfrac{a}{1-r} \\\\
& =\dfrac{81 / 4}{1-\dfrac{2}{3}} \\\\
& =\dfrac{81}{4} \times 3 \\\\
& =\dfrac{243}{4}
\end{aligned}$
- The sum of an infinite geometric progression is $12$ and its first term is $3$ .
Fin the first four terms of the G.P.
SOLUTION
$\begin{aligned}
\text { In a G.P., } & \\\\
S=12, a&=3 \\\\
\therefore\ \dfrac{3}{1-r}&=12 \\\\
1-r&=\dfrac{1}{4} \\\\
r&=\dfrac{3}{4} \\\\
\therefore u_{1}=a&=3 \\\\
u_{2}=a r&=\dfrac{9}{4} \\\\
u_{3}=a r^{2}&=\dfrac{27}{16} \\\\
u_{4}=a r^{3}&=\dfrac{81}{64}
\end{aligned}$
- In a G.P. the ratio of the sum of the first three terms to the sum to infinity is
$19: 27$. Find the common ratio.
SOLUTION Let the first term be $a$ and the common ratio be $r$ for the given G.P.
By the problem,
$\begin{aligned}
&\\
\dfrac{S_{3}}{S}&=\dfrac{19}{27} \\\\
\dfrac{\dfrac{a\left(1-r^{3}\right)}{1-r}}{\dfrac{a}{1-r}}&=\dfrac{19}{27} \\\\
1-r^{3}&=\dfrac{19}{27} \\\\
r^{3}&=\dfrac{8}{27}\\\\
\therefore r&=\dfrac{2}{3}
\end{aligned}$
- The second term of a G.P. is $2$ and its sum to infinity is $9$ .
Find the sum of the first $4$ terms of the two possible geometric progressions.
SOLUTION
$\begin{aligned}
\text { In a G.P., }\quad & \\\\
u_{2}&=2 \\\\
a r&=2 \\\\
a&=\dfrac{2}{r} \\\\
S&=9 \\\\
\dfrac{a}{1-r}&= 9 \\\\
\dfrac{\dfrac{2}{r}}{1-r}&=9 \\\\
\dfrac{2}{r}&=9-9 r \\\\
9 r^{2}-9 r+2&=0\\\\
(3 r-1)(3 r-2)&=0 \\\\
r=\dfrac{1}{3} & \text { or } r=\dfrac{2}{3} \\\\
\end{aligned}$
$\begin{aligned}
\text { When } r&=\dfrac{1}{3}, \\\\
a&=\dfrac{2}{1 / 3}=6 \\\\
\therefore S_4&=\dfrac{a(1-r^4)}{1-r} \\\\
& = \dfrac{6 (1-(1/3)^4)}{1-1/3}\\\\
& = \dfrac{6 (1-\dfrac{1}{81})}{\dfrac{2}{3}}\\\\
& = \dfrac{6 \times\dfrac{80}{81}}{\dfrac{2}{3}}\\\\
&=\dfrac{80}{9}\\\\
\text { When } r&=\dfrac{2}{3}, \\\\
a&=\dfrac{2}{2 / 3}=3 \\\\
\therefore S_4&=\dfrac{a(1-r^4)}{1-r} \\\\
& = \dfrac{3(1-(2/3)^4)}{1-2/3}\\\\
& = \dfrac{3 (1-\dfrac{16}{81})}{\dfrac{1}{3}}\\\\
& = \dfrac{3 \times\dfrac{65}{81}}{\dfrac{1}{3}}\\\\
&=\dfrac{65}{9}
\end{aligned}$
- The sum of the first 3 terms of a G.P. is $27$ and the sum of the fourth, fifth and
sixth terms is $-1$. Find the common ratio and the sum to infinity of the G.P.
SOLUTION
$\begin{aligned}
\text { In a G.P., }& \\\\
u_{1}+u_{2}+u_{3}&=27 \\\\
a+a r+a r^{2}&=27 \ldots(1) \\\\
u_{4}+u_{5}+u_{6}&=-1 \\\\
a r^{3}+a r^{4}+a r^{5}&=-1 \cdots(2) \\\\
\text {By }(2) \div(1),& \\\\
\dfrac{a r^{3}+a r^{4}+a r^{5}}{a+a r+a r^{2}}&=-\dfrac{1}{27} \\\\
\dfrac{r^{3}\left(a+a r+a r^{2}\right)}{\left(a+a r+a r^{2}\right)}&=-\dfrac{1}{27} \\\\
\therefore\ r^{3}&=-\dfrac{1}{27}\\\\
r&=-\dfrac{1}{3}\\\\
\end{aligned}$
$\begin{aligned}
&\text{Substitue } r=-\dfrac{1}{3} \text{ in equation } (1),\\\\
&a+a \left(-\dfrac{1}{3}\right) +a \left(-\dfrac{1}{3}\right)^2 =27 \\\\
&a - \dfrac{a}{3} + \dfrac{a}{9} = 27\\\\
&\dfrac{7a}{9}= 27\\\\
&a=\dfrac{243}{7}\\\\
\end{aligned}$
$\begin{aligned}
S&=\dfrac{a}{1-r}\\\\
&=\dfrac{\dfrac{243}{7}}{1+\dfrac{1}{3}}\\\\
&=\dfrac{729}{28}
\end{aligned}$
- Given that $x+18, x+4$ and $x-8$ are the first three terms of a G.P.,
find the value of $x$. Hence find
(a) the common ratio
(b) the fifth term
(c) the sum to infinity.
SOLUTION
$\begin{aligned}
&x+18, x+4, x-8 \text { are in G.P. } \\\\
&\dfrac{x+4}{x+18}=\dfrac{x-8}{x+4} \\\\
&\therefore\ x^{2}+8 x+16 = x^{2}+10 x-144 \\\\
\end{aligned}$
$\begin{aligned}
2x &= 160 \\\\
x &=80 \\\\
\therefore\ u_{1}&=98 \\\\
u_{2}&= 84 \\\\
u_{3}&= 72 \\\\
\therefore\ a&=98 \\\\
r&=\dfrac{84}{98}\\\\
&=\dfrac{6}{7} \\\\
\end{aligned}$
$\begin{aligned}
\therefore\ u_5 &= ar^4\\\\
&= 98 \times\left(\dfrac{6}{7}\right)^{4}\\\\
&=\dfrac{2592}{49}\\\\
S &= \dfrac{a}{1-r}\\\\
&= \dfrac{98}{1-\dfrac{6}{7}}\\\\
&=686
\end{aligned}$
- Given that $2 x-14, x-4$ and $\dfrac{1}{2} x$ are successive terms of a sequence.
(a) find the value of $x$ when the sequence is (i) an A.P. (ii) a G.P.
(b) If $2 x-14$ is the $3^{\text {rd }}$ term of a G.P. with infinite terms,
find
(i) the common ratio
(ii) the sum to infinity.
SOLUTION
$2 x-14, x-4, \dfrac{1}{2} x$ are successive terms of a sequence.
When the sequence is an A.P.,
$\begin{aligned}
&\\
x-4-(2 x-14)&=\dfrac{1}{2} x-(x-4) \\\\
-x+10&=-\dfrac{1}{2} x+4 \\\\
\dfrac{1}{2} x&=6 \\\\
x&=12\\\\
\end{aligned}$
When the sequence is G.P.,
$\begin{aligned}
&\\
\dfrac{x-4}{2 x-14}&=\dfrac{\dfrac{1}{2} x}{x-4}\\\\
x^{2}-8 x+16&=x^{2}-7 x \\\\
\therefore\ x&=16 \\\\
u_{3}&=a r^{2}\\\\
&=2 x-4\\\\&=28 \\\\
u_{4}&=a r^{3}\\\\
&=x-4\\\\&=14 \\\\
\therefore\ r&=\dfrac{u_{4}}{u_{3}}\\\\
&=\dfrac{14}{28}\\\\
&=\dfrac{1}{2} \\\\
\therefore\ a\left(\dfrac{1}{2}\right)^{2}&=28 \\\\
S&=\dfrac{a}{1-r}\\\\
&=\dfrac{112}{1-\dfrac{1}{2}}\\\\
&=224
\end{aligned}$
- A G.P. has a first term $18$ and a sum to infinity of $30$.
Each of the terms in the progression is squared to form a new G.P.
Find the sum to infinity of the new progression.
SOLUTION
Let the first termbe $a$ and the common ratio be $r$ for the given G.P.
$\begin{aligned}
&\\
a&=18 \\\\
S&=30 \\\\
\dfrac{a}{1-r}&=30 \\\\
\dfrac{18}{1-r}&=30 \\\\
1-r&=\dfrac{3}{5} \\\\
\therefore\ r&=\dfrac{2}{5}\\\\
\end{aligned}$
Each tems of the progression is squared, the new sequence formed is
$a^{2}, a^{2} r^{2}, a^{2} r^4, a^{2}r^6, \ldots\\\\ $
Let the first term be $A$ and the common ratio be $R$ for the new G.P.,
$\begin{aligned}
&\\
\therefore A &=a^{2}=18^{2} \\\\
R &=\dfrac{a^{2} r^{2}}{a^{2}}\\\\
&=r^{2}\\\\
&=\dfrac{4}{25}\\\\
\end{aligned}$
Let the sum to infinity be $S$.
$\begin{aligned}
&\\
\therefore\ S &=\dfrac{A}{1-R} \\\\
&=\dfrac{18^{2}}{1-\dfrac{4}{25}} \\\\
&=\dfrac{2700}{7}
\end{aligned}$
- The first term of a G.P is $a$ and the common ratio is $r$.
Given that $a+96 r=0$ and that the sum to infinity is $32$ , find the $8^{\text {th }}$ term.
- An infinite geometric series has a finite sum of $256$ .
The sum of the first $3$ terms is $224$. What is the value of the $3^{\text {rd }}$ term?
SOLUTION
$\begin{aligned}
S&=256 \\\\
\dfrac{a}{1-r}&=256 \ldots(1) \\\\
S_{3}&=224 \\\\
\dfrac{a\left(1-r^{3}\right)}{1-r}&=224 \cdots(2) \\\\
\text{By } & (2) \div(1), \\\\
\dfrac{\dfrac{a\left(1-r^{3}\right)}{1-r}}{\dfrac{a}{1-r}}&=\dfrac{224}{256} \\
1-r^{3}&=\dfrac{7}{8}\\\\
r^3&=\dfrac{1}{8}\\\\
r&=\dfrac{1}{2}\\\\
\text{Substitue }& r=\dfrac{1}{2} \text{ in equation (1).}\\\\
\dfrac{a}{1-\dfrac{1}{2}}&=256\\\\
a&=128\\\\
\therefore\ u_3&=ar^2=128\left(\dfrac{1}{2}\right)^2=32
\end{aligned}$
- In a G.P., whose first term is positive, the sum of the first and third terms
$\dfrac{13}{9}$ and the product of second and fourth terms is
$\dfrac{16}{81}$. Find the common ratio and the sum to infinity.
SOLUTION
$\begin{aligned}
&\text { In a G.P., } \\\\
&a>0 \\\\
&u_{1}+u_{3}=\dfrac{13}{9} \\\\
&a+ar^{2}=\dfrac{13}{9} \cdots(1) \\\\
&u_{2} \cdot u_{4}=\dfrac{16}{81} \\\\
&ar \cdot a r^{3}=\dfrac{16}{81} \\\\
&\left(a r^{2}\right)^{2}=\dfrac{16}{8}\\\\
&a r^{2}=\dfrac{4}{9} \quad(\because a>0) \quad \ldots(2)\\\\
\end{aligned}$
Substitue $a r^{2}=\dfrac{4}{9}$ in equation (1),
$\begin{aligned}
&\\
&a+\dfrac{4}{9}=\dfrac{13}{9} \\\\
&a=1\\\\
\end{aligned}$
Substitue $a=1$ in equation (2),
$\begin{aligned}
&\\
&r^{2}=\dfrac{4}{9} \\\\
&r=\pm \dfrac{2}{3}\\\\
\end{aligned}$
When $a=1, r=-\dfrac{2}{3}$
$\begin{aligned}
&\\
S &=\dfrac{a}{1-r} \\\\
&=\dfrac{1}{1+\dfrac{2}{3}} \\\\
&=\dfrac{3}{5}\\\\
\end{aligned}$
When $a=1, r=\dfrac{2}{3}$
$\begin{aligned}
&\\
S &=\dfrac{a}{1-r} \\\\
&=\dfrac{1}{1-\dfrac{2}{3}} \\\\
&=3
\end{aligned}$
- In a G.P. the sum of the first term and the fourth term is $144$ ,
and the first term exceeds the seventh term by $126$ .
Find the first term, the common ratio and the sum to infinity of the G.P.
SOLUTION
$\begin{aligned}
&\text { In a G.P, } \\\\
&u_{1}+u_{4}=144 \\\\
&a+a r^{3}=144 \cdots(1) \\\\
&u_{1}-u_{7}=126 \\\\
&a-a r^{6}=126 \cdots(2) \\\\
&\text{By } (2) \div(1), \\\\
&\dfrac{a-a r^{6}}{a+a r^{3}}=\dfrac{126}{144}\\\\
&\dfrac{a\left(1-r^{6}\right)}{a\left(1+r^{3}\right)}=\dfrac{7}{8} \\\\
&\dfrac{\left(1+r^{3}\right)\left(1-r^{3}\right)}{1+r^{3}}=\dfrac{7}{8} \\\\
&1-r^{3}=\dfrac{7}{8} \\\\
&r^{3}=\dfrac{1}{8} \\\\
&r=\dfrac{1}{2}\\\\
\end{aligned}$
Substitue $r=\dfrac{1}{2}$ in equation (1).
$\begin{aligned}
&\\
a+\dfrac{a}{8}&=144\\\\
\dfrac{9 a}{8} &=144 \\\\
a &=128 \\\\
S &=\dfrac{a}{1-r} \\\\
&=\dfrac{128}{1-\dfrac{1}{2}} \\\\
&=256
\end{aligned}$
- The first term of a G.P. is $3$ and the common ratio is $\dfrac{2}{5}$.
Find the sum to infinity, and the least value of $n$ for which the sum to $n$
terms differ from the sum to infinity by less than $0.04$.
SOLUTION
$\begin{aligned}
\text{In a G.P, } \\\\
a &=3, r=\dfrac{2}{5} \\\\
S &=\dfrac{a}{1-r} \\\\
&=\dfrac{3}{1-\dfrac{2}{5}} \\\\
&=5 \\\\
S_{n} &=\dfrac{a\left(1-r^{n}\right)}{1-r} \\\\
&=5\left[1-\left(\dfrac{2}{5}\right)^{n}\right]\\\\
\end{aligned}$
By the problem,
$\begin{aligned}
&\\
S-S_{n} & < 0.04 \\\\
5-5\left[1-\left(\dfrac{2}{5}\right)^{n}\right] & < 0.04 \\\\
5\left(\dfrac{2}{5}\right)^{n} & < 0.04 \\\\
\left(\dfrac{2}{5}\right)^{n} & <0.008 \\\\
\left(\dfrac{5}{2}\right)^{n} & <125 \\\\
(2.5)^{n} &>125\\\\
n&>\log _{2.5} 125\\\\
n&>\dfrac{\ln 125}{\ln 2.5}\\\\
n&>5.269\\\\
\end{aligned}$
$\therefore$ The least value of $n=6$.
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