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- Question 1 and 2 Exercise 4.1
- nd infinite sequences\\ $2,4,6,8, \ldots ,50$ ====Solution==== It is finite sequence whose last term is $50... and infinite sequences. $1,0,1,0,1, \ldots$. ====Solution==== It is infinite sequence, the last term may be... nfinite sequences: $...,-4,0,4,8, \ldots, 60$ ====Solution==== This is infinite sequence. =====Question 1(i... }{9},-\dfrac{1}{27}, \ldots,-\dfrac{1}{2187}$ ====Solution==== This finite sequence. =====Question 2(i)====
- Question 1 Exercise 4.5
- i)===== Compute the sum $3+6+12+\ldots+3.2^9$ ====Solution==== In the given geometric series: $a_1=3, \quad ... Compute the sum $8+4+2+\ldots+\dfrac{1}{16}$ ====Solution==== In the give geometric series $$a_1=8, \quad r... = Compute the sum $2^4+2^5+2^6+\ldots+2^{10}$ ====Solution==== In the give geometric series.\\ $$a_1=2^4, \q... sum $\dfrac{8}{5},-1, \dfrac{5}{8}, \ldots$, ====Solution==== Here $$a_1=\dfrac{8}{5}$$ \begin{align}r&=\d
- Question 3 and 4 Exercise 4.1
- rac{2}{3} \dfrac{3}{4}, \dfrac{4}{5}, \ldots$ ====Solution==== We can reform the given sequence to pick the ... gested by the pattern. $2,-4,6,-8,10, \ldots$ ====Solution==== We can reform the given sequence to pick the ... suggested by the pattern. $1,-1,1,-1, \ldots$ ====Solution==== We can reform the give sequence to pick the p... efined recursively. $a_1=3$, $a_{n+1}=5-a_n$. ====Solution==== Given $$a_1=3, a_{n+1}=5-a_n.$$ For $n=1$ \be
- Question 12 & 13 Exercise 4.2
- ry during his twenty first year of work? GOOD ====Solution==== Suppose $a_1$ represents salary of worker at ... e arithmetic mean between $12$ and $18$. GOOD ====Solution==== Here $a=12, b=18$.\\ Let say $A$ be arithmeti... an between $\dfrac{1}{3}$ and $\dfrac{1}{4}$. ====Solution==== Here $a=\dfrac{1}{3}, b=\dfrac{1}{4}$,\\ Let ... d the arithmetic mean between $-6,-216$. GOOD ====Solution==== Here $a=-6, b=-216$.\\ Let $A$ be arithmetic
- Question 5 Exercise 4.1
- eries in expanded form, $\sum_{j=1}^6(2 j-3)$ ====Solution==== \begin{align}\sum_{j=1}^6(2 j-3)&=(2.1-3)+(2.... n expanded form, $\sum_{k=1}^5(-1)^k 2^{k-1}$ ====Solution==== \begin{align}\sum_{k=1}^5(-1)^k 2^{k-1}& =(-1... ed form, $\sum_{j=1}^{\infty} \dfrac{1}{2^j}$ ====Solution==== \begin{align}\sum_{j=1}^{\infty} \dfrac{1}{2^... um_{k=0}^{\infty}\left(\dfrac{3}{2}\right)^k$ ====Solution==== \begin{align}\sum_{k=0}^{\infty}\left(\dfrac{
- Question 2 Exercise 4.3
- one that is missing: $a_1=2, n=17, d=3$. GOOD ====Solution==== Given: $a_1=2, n=17, d=3$ \\ We need to find ... that are missing $a_1=-40, S_{21}=210$. GOOD ====Solution==== Given: $a_1=-40$ and $S_{21}=210$.\\ So we ha... that are missing $a_1=-7, d=8, S_n=225$. GOOD ====Solution==== Given: $a_1=-7, d=8, S_n=225$, we have to fin... ne that are missing: $a_n=4, S_{15}=30$. GOOD ====Solution==== Given: $a_n=4, S_{15}=30$.\\ Thus we have $n=
- Question 1 Exercise 4.4
- ic ric sequence given that $a_1=5, \quad r=3$ ====Solution==== The gcometric sequence is $a_1, a_1 r, a_1 r^... nce given that $a_1=8, \quad r=-\dfrac{1}{2}$ ====Solution==== The geomerric sequence is $a_1, a_1 r, a_1 r^... t $a_1=-\dfrac{9}{16}, \quad r=-\dfrac{2}{3}$ ====Solution==== The geometric sequence is $a_1, a_1 r, a_1 r^... hat $a_1=\dfrac{x}{y}, \quad r=-\dfrac{y}{x}$ ====Solution==== The geometric sequence is, $a_1, a_1 r, a_1 r
- Question 8 Exercise 4.4
- Find the geometric mean of $3.14$ and $2.71$ ====Solution==== Here $a=3.14$ and $b=2.71$\\ then $$G= \pm \s... == Find the geometric mean of $-6$ and $-216$ ====Solution==== Here $a=-6$ and $b=-216$ then\\ \begin{align}... == Find the geometric mean of $x+y$ and $x-y$ ====Solution==== Here $a=x+y$ and $b=x-y$\\ then $$G= \pm \sqr... ometric mean of $\sqrt{2}+3$ and $\sqrt{2}-3$ ====Solution==== Here $a=\sqrt{2}+3$ and \begin{align}b&=\sqrt
- Question 4 Exercise 4.5
- decimal to common fraction $0 . \overline{8}$ ====Solution==== We can write $$0 . \overline{8}=0.888888 \ldo... ecimal to common fraction $1 . \overline{63}$ ====Solution==== Since \begin{align}1 . \overline{63}&=1+0.63+... ecimal to common fraction $2 . \overline{15}$ ====Solution==== Since \begin{align}2 . \overline{15}&=2+0.15+... cimal to common fraction $0 . \overline{123}$ ====Solution==== Since $$0 . \overline{123}=0.123+0.000123 +0.
- Question 6 Exercise 4.1
- using its general recursive definition. GOOD ====Solution==== For $n=5$, we have Pascal sequence as follow... 6$ by using its general recursive definition. ====Solution==== As we know the general definition of Pascal s... $ by using its general recursive definition. =====Solution===== As we know the general definition of Pascal
- Question 2 Exercise 4.5
- re missing $a_1=1, \quad r=-2, \quad a_n=64$. ====Solution==== We first find $n$ and then $S_n$\\ We know $a... that are missing $r=\dfrac{1}{2}, a_9=1, n=9$ ====Solution==== We first find $a_1$ and then $S_9$.\\ We kno... nes that are missing $r=-2, S_n=-63, a_n=-96$ ====Solution==== We know that\\ \begin{align}S_n&=\dfrac{a_1(r
- Question 1 and 2 Exercise 4.2
- the arithmetic sequence $2,5,8, \ldots$ GOOD ====Solution==== Here $a_1=2$, $d=5-2=3$ and $n=15$. We know t... and the 21st is 108. Find the 7th term. GOOD ====Solution==== Since $a_1=8$ and $a_{21}=108$. We know that
- Question 3 and 4 Exercise 4.2
- arithmetic progression $6,9,12, \ldots, 78$. ====Solution==== Here $a_1=6$ and $d=9-6=3$ and $a_n=78$.\\ We... ithmetic progression. Also find its 7th term. ====Solution==== Given that $$a_n=2 n+7. --- (1)$$ Then \begin
- Question 5 and 6 Exercise 4.2
- , \ldots$$ is an A.P. Also find its nth term. ====Solution==== We first find $n$th term. Each term of the se... k-4$ are in A.P. Also find the sequence. GOOD ====Solution==== Since the given terms are in A.P, \begin{alig
- Question 14 Exercise 4.2
- three arithmetic means between 6 and 41. GOOD ====Solution==== Let $A_1, A_2, A_3$ be three arithmetic means... four arithmetic means between 17 and 32. GOOD ====Solution==== Let $A_1, A_2, A_3, A_4$ be four arithmetic m