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- Question 20 and 21, Exercise 4.4
- $ and $a_5=48$. Assume $r$ be common difference, then by general formula for nth term, we have $$ a_n=a... $ and $a_5=48$. Assume $r$ be common difference, then by general formula for nth term, we have $$ a_n=a... = \pm 2. \end{align*} Thus, if $a_1=3$ and $r=2$, then \begin{align*} & a_2=a_1 r= (3)(2) = 6 \\ & a_3=a... (3)(2)^3=24. \end{align*} If $a_1=3$ and $r=-2$, then \begin{align*} & a_2=a_1 r= (3)(-2) = -6 \\ & a_3
- Question 5, 6 and 7, Exercise 4.4
- $r=-2$. Use the formula $$a_{n}=a_{1} r^{n-1}.$$ Then \begin{align*} & a_{2}=a_{1} r=(3)(-2)= -6 \\ & a... 1}{3}$. Use the formula $$a_{n}=a_{1} r^{n-1}.$$ Then \begin{align*} & a_{2}=a_{1} r=(27)\left(-\frac{1... 1}{2}$. Use the formula $$a_{n}=a_{1} r^{n-1}.$$ Then \begin{align*} & a_{2}=a_{1} r=(12)\left(\frac{1}
- Question 14, 15 and 16, Exercise 4.7
- nsider $T_n$ represents the $n$th term of series, then $$ T_{n} = n+1. $$ Taking summation \begin{alig... nsider $T_k$ represents the $k$th term of series, then $$ T_{k} = k^{2} + 2k $$ Taking summation, we ha... nsider $T_k$ represents the $k$th term of series, then $$ T_{k} = 3k^{2} + 2k + 1 $$ Taking summation,
- Question 1 and 2, Exercise 4.1
- }=3 n+1$$ ** Solution. ** Given $$a_{n}=3 n+1$$ Then \begin{align*} a_1 &= 3(1) + 1 = 3 + 1 = 4\\ a_2 ... =3 n-1$$ ** Solution. ** Given: $$a_{n}=3 n-1.$$ Then \begin{align*} a_1 &= 3(1) - 1 = 3 - 1 = 2\\ a_2
- Question 5 and 6, Exercise 4.1
- }-2 n$ ** Solution. ** Given $$a_n = n^2 - 2n.$$ Then \begin{align*} a_1 &= (1)^2 - 2(1) = 1 - 2 = -1\\... tion. ** Given $$a_n = \frac{n^2 - 1}{n^2 + 1}.$$ Then \begin{align*} a_1 &= \frac{1^2 - 1}{1^2 + 1} = \
- Question 15 and 16, Exercise 4.1
- ** Solution. ** Given: $$a_n = 4n^2(11n + 31).$$ Then \begin{align*} a_{22} &= 4 \cdot 22^2 \cdot (11 \... Given: $$a_n = \left(1 + \frac{1}{n}\right)^2.$$ Then \begin{align*} a_{20} &= \left(1 + \frac{1}{20}\r
- Question 17 and 18, Exercise 4.1
- 43}$ ** Solution. ** Given: $$a_n = \log 10^n.$$ Then \begin{align*} a_{43} &= \log 10^{43} \\ &= 43 \c... _{67}$ ** Solution. ** Given: $$a_n = \ln e^n.$$ Then \begin{align*} a_{67} &= \ln e^{67} \\ &= 67 \cdo
- Question 16 and 17, Exercise 4.2
- $ be two arithmetic means between $5$ and $17$.\\ Then $5$, $A_1$, $A_2$, $17$ are in A.P.\\ Here $a_1=5... be thre arithmetic means between $2$ and $-18$.\\ Then $2$, $A_1$, $A_2$, $A_3$, $-18$ are in A.P.\\ Her
- Question 3 and 4, Exercise 4.3
- \\ Let $S_n$ represents sum of arithmetic series. Then \begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \impli... \\ Let $S_n$ represents sum of arithmetic series. Then \begin{align} S_n&=\frac{n}{2}[2a_1+(n-1)d] \\ \i
- Question 5 and 6, Exercise 4.3
- \\ Let $S_n$ represents sum of arithmetic series. Then \begin{align} S_n&=\frac{n}{2}[2a_1+(n-1)d] \\ \i... \\ Let $S_n$ represents sum of arithmetic series. Then \begin{align} S_n&=\frac{n}{2}[2a_1+(n-1)d] \\ \i
- Question 7 and 8, Exercise 4.3
- et $S_n$ represents sum of the arithmetic series. Then \begin{align} S_n&=\frac{n}{2}[2a_1+(n-1)d] \\ \i... et $S_n$ represents sum of the arithmetic series. Then \begin{align} S_n&=\frac{n}{2}[2a_1+(n-1)d] \\ \i
- Question 9 and 10, Exercise 4.3
- et $S_n$ represents sum of the arithmetic series. Then \begin{align} S_n&=\frac{n}{2}[2a_1+(n-1)d] \\ \i... et $S_n$ represents sum of the arithmetic series. Then \begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \impli
- Question 11 and 12, Exercise 4.3
- olution. ** Given $a_{1}=3$, $a_{n}=-38$, $n=8$, then \begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \impli... lution. ** Given $a_{1}=85$, $n=21$, $a_{n}=25$, then \begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \impli
- Question 13 and 14, Exercise 4.3
- olution. ** Given $a_{1}=34$, $n=9$, $a_{n}=2$., then \begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \impli... . ** Given: $a_{1}=5$, $d=\frac{1}{2}$, $n=13$, then \begin{align} S_n&=\frac{n}{2}[2a_1+(n-1)d] \\ \i
- Question 22 and 23, Exercise 4.4
- \frac{1}{4}$. Assume $r$ to be the common ratio. Then, by the general formula for the $n$th term, we ha... nd $a_3=75$. Assume $r$ to be the common ratio. Then, by the general formula for the $n$th term, we ha