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- Question 1, Exercise 4.2
- ic sequence with $a_{1}=4, d=3$ ** Solution. ** Given: $a_1= 4$, $d=3$.\\ The general term of an arithm... ic sequence with $a_1=7$, $d=5$ ** Solution. ** Given: $a_1= 7$, $d=5$.\\ The general term of an arithm... c sequence. $a_{1}=16$, $d=-2$. ** Solution. ** Given: $a_1= 16$, $d=-2$.\\ We have $$a_n = a_1 + (n - ... tic sequence. $a_1=38$, $d=-4$. ** Solution. ** Given: $a_1= 38$, $d=-4$.\\ We have $$a_n = a_1 + (n -
- Question 17, 18 and 19, Exercise 4.3
- ic series. $6+12+18+\ldots+96$. ** Solution. ** Given arithmetic series: $$6+12+18+\ldots+96.$$ So, $a... \times 102\\ &=1224. \end{align} Hence the sum of given series is $1224$. =====Question 18===== Find sum ... etic series. $34+30+26+\ldots+2$ ** Solution. ** Given arithmetic series: $$34+30+26+\ldots+2.$$ So, $a... imes 36\\ &=162. \end{align} Hence the sum of the given series is $162$. =====Question 19===== Find
- Question 1 and 2, Exercise 4.4
- n ratio. $5,20,100,500, \ldots$ ** Solution. ** Given sequence is $5, 20, 100, 500, \ldots $.\\ A seque... {5} = 4\neq \frac{100}{20} = 5.\end{align*} Hence given sequence is not geometric. GOOD **Alternative Method** Given sequence is $5, 20, 100, 500, \ldots $.\\ Suppose... two consective terms has different ratio.\\ Hence given sequence is not geometric. GOOD =====Question
- Question 5 and 6, Exercise 4.2
- n. ** The nth term of the arithmetic sequence is given as $$a_n=a_1+(n-1)d$$ Given \begin{align*} & a_{17} = -40 \\ \implies &a_1 + 16d = -40 \quad \cdots (1) ... the first term and 87 th term. ** Solution. ** Given: $a_5 = 19$ and $a_{11} = 43$. The nth term of the arithmetic sequence is given as $$a_n = a_1 + (n-1)d$$ Given \begin{align*} &
- Question 1 and 2, Exercise 4.1
- Question 1===== The $n$th term of the sequence is given, find the first 4 terms; the 10th term, $a_{10}$ ... term: $a_{15}$. $$a_{n}=3 n+1$$ ** Solution. ** Given $$a_{n}=3 n+1$$ Then \begin{align*} a_1 &= 3(1) ... Question 2===== The $n$th term of the sequence is given, find the first 4 terms; the 10th term, $a_{10}$ ... term, $a_{15}$. $$a_{n}=3 n-1$$ ** Solution. ** Given: $$a_{n}=3 n-1.$$ Then \begin{align*} a_1 &= 3(1)
- Question 5 and 6, Exercise 4.1
- Question 5===== The $n$th term of the sequence is given, find the first 4 terms; the 10th term, $a_{10}$ ... term: $a_{15}$.$a_{n}=n^{2}-2 n$ ** Solution. ** Given $$a_n = n^2 - 2n.$$ Then \begin{align*} a_1 &= (1... Question 6===== The $n$th term of the sequence is given, find the first 4 terms; the 10th term, $a_{10}$ ... $a_{n}=\frac{n^{2}-1}{n^{2}+1}$ ** Solution. ** Given $$a_n = \frac{n^2 - 1}{n^2 + 1}.$$ Then \begin{al
- Question 3 and 4, Exercise 4.2
- equence $0.07,0.12,0.7, \ldots$ ** Solution. ** Given $$0.07,0.12,0.7, \ldots$$ From this, we have $a_1... irst four terms of the sequence. ** Solution. ** Given: $a_3 = 14$ and $a_9 = -1$. The nth term of the arithmetic sequence is given as $$a_n = a_1 + (n-1)d$$ Given \begin{align*} & a_3 = 14 \\ \implies & a_1 + 2d = 14 \quad \cdots (1) \
- Question 7 and 8, Exercise 4.2
- ence $-6,-2,2, \ldots$ is $70$? ** Solution. ** Given $-6,-2,2, \ldots$ is an arithmetic sequence. Her... n=?$. The nth term of the arithmetic sequence is given as $$a_n=a_1+(n-1)d.$$ This gives \begin{align*} ... , \ldots$ is $-\dfrac{105}{2}$? ** Solution. ** Given: $\dfrac{5}{2}, \dfrac{3}{2}, \dfrac{1}{2}, \ldot... n=?$. The nth term of the arithmetic sequence is given as $$a_n = a_1 + (n-1)d.$$ This gives \begin{alig
- Question 1 and 2, Exercise 4.3
- series: $4+7+10+13+16+19+22+25$ ** Solution. ** Given: $4+7+10+13+16+19+22+25$. As the given series is arithmetic series with $a_1=4$, $d=7-4=3$, $n=8$, so \... \ &=4[8+7\times 3] = 116 \end{align} Hence sum of given series is 116. GOOD =====Question 2===== Find the... $a_{1}=2$, $a_{n}=200$, $n=100$ ** Solution. ** Given $a_{1}=2$, $a_{n}=200$, $n=100$.\\ Let $S_n$ repr
- Question 3 and 4, Exercise 4.4
- tan. =====Question 3===== Determine whether the given sequence is geometric. If so, find the common rat... c{27}{8}, \frac{81}{16}, \ldots$ ** Solution. ** Given sequence is \(\frac{3}{2}, \frac{9}{4}, \frac{27}... ans two consecutive terms has same ratio.\\ Hence given sequence is geometric and common ratio \(r = \fra... mon ratio. $7,14,21,28, \ldots$ ** Solution. ** Given sequence is $7, 14, 21, 28, \ldots$ \\ Suppose \b
- Question 8 and 9, Exercise 4.4
- c sequence: $$90,30,10 \ldots$$ ** Solution. ** Given sequence is geometric with $a_1=90$ and $r=\dfrac... 1}{3}$.\\ General term of the geometric series is given as $$a_{n}=a_{1} r^{n-1}.$$ Thus \begin{align*} ... ic sequence: $$2,6,18 \ldots$$ ** Solution. ** Given sequence is geometric with \(a_1=2\) and \(r=\fra... }{1}\).\\ General term of the geometric series is given as $a_{n}=a_{1} r^{n-1}.$ Thus \begin{align*} &
- Question 10 and 11, Exercise 4.4
- sequence: $$20,30,45 \ldots$$ ** Solution. ** Given sequence is geometric with \(a_1=20\) and \(r=\fr... }{2}\).\\ General term of the geometric series is given as $$a_{n}=a_{1} r^{n-1}.$$ Thus \begin{align*} ... equence: $$729,243,81,\ldots$$ ** Solution. ** Given sequence is geometric with \(a_1=729\) and \(r=\f... }{3}\).\\ General term of the geometric series is given as $a_{n}=a_{1} r^{n-1}.$ Thus \begin{align*} &
- Question 12 and 13, Exercise 4.4
- ac{1}{9}, \frac{1}{3}, \ldots$$ ** Solution. ** Given sequence is geometric with \(a_1=\frac{1}{27}\) a... }}=3\).\\ General term of the geometric series is given as $a_{n}=a_{1} r^{n-1}.$ Thus \begin{align*} & ... 1}{4}, \frac{1}{2},-1, \ldots$$ ** Solution. ** Given sequence is geometric with \(a_1=\frac{1}{4}\) an... }}=2\).\\ General term of the geometric series is given as $a_{n}=a_{1} r^{n-1}.$ Thus \begin{align*} &
- Question 14 and 15, Exercise 4.4
- equence if $a_{1}=4, n=3, r=5$. ** Solution. ** Given $a_{1}=4, n=3, r=5$.\\ General term of the geometric series is given by $$a_{n}=a_{1} r^{n-1}.$$ Thus \begin{align*} a... equence if $a_{1}=2, n=5, r=2$. ** Solution. ** Given $a_{1}=2$, $n=5$, $r=2$.\\ General term of the geometric series is given by $a_{n}=a_{1} r^{n-1}.$ Thus \begin{align*} a_5
- Question 16 and 17, Exercise 4.4
- sequence if $a_{1}=7, n=4, r=2$ ** Solution. ** Given $a_{1}=7$, $n=4$, $r=2$.\\ General term of the geometric series is given by $a_{n}=a_{1} r^{n-1}.$ Thus \begin{align*} a_4... a_{1}=243, n=5, r=-\frac{1}{3}$ ** Solution. ** Given $a_{1}=243$, $n=5$, $r=-\frac{1}{3}$.\\ General term of the geometric series is given by $a_{n}=a_{1} r^{n-1}.$ Thus \begin{align*} a_5