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- Question 5 and 6, Exercise 4.8
- _{n-1})-T_{n}. \\ \implies 0=&3+(1+2+4+8+\ldots \text { up to } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =3+(1+2+4+8+\ldots \text { up to }(n-1) \text { terms }) \\ & =3+\frac{1(2^{n-1}-1)}{2-1} \quad\left(\because S_{n}=\frac{a(r^n-
- Question 1 and 2, Exercise 4.8
- ight)-T_{n}. \\ \implies 0=&3+(4+6+8+10+\ldots \text { up to } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =3+(4+6+8+10+\ldots \text { up to }(n-1) \text { terms }) \\ & =3+\frac{n-1}{2}[2(4)+(n-1-1)(2)] \quad\left(\because S_{n}=\frac{
- Question 3 and 4, Exercise 4.8
- -1})-T_{n}. \\ \implies 0=&1+(3+9+27+81+\ldots \text { up to } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =1+(3+9+27+81+\ldots \text { up to }(n-1) \text { terms }) \\ & =1+\frac{3(3^{n-1}-1)}{3-1} \quad\left(\because S_{n}=\frac{a(r^n
- Question 13, Exercise 4.2
- ** Here $a=7$ and $b=17$.\\ Now \begin{align*} \text{A.M.} &= \frac{a + b}{2}\\ &= \frac{7 + 17}{2} \\... rt{2}$ and $b=7-3\sqrt{2}$.\\ Now \begin{align*} \text{A.M.} &= \frac{a + b}{2}\\ &= \frac{(3 + 3 \sqrt{... \sqrt{5}$ and $b=\sqrt{5}$.\\ Now \begin{align*} \text{A.M.} &= \frac{a + b}{2}\\ &= \frac{7 \sqrt{5} + ... Here $a=2y+5$ and $b=5y+3$.\\ Now \begin{align*} \text{A.M.} &= \frac{a + b}{2}\\ &= \frac{(2y + 5) + (5
- Question 3 & 4, Exercise 4.6
- {1}{18}, \frac{1}{13}, \frac{1}{8}, \ldots \quad \text{ is in H.P.} \\ &18, 13, 8, \ldots \quad \text{ is in A.P.} \end{align*} Here $a_1 = 18$, $d = 13 - 18 ... c{1}{4}, \frac{1}{9}, \frac{1}{14}, \ldots \quad \text{ is in H.P.} \\ &4, 9, 14, \ldots \quad \text{ is in A.P.} \end{align*} Here $a_1 = 4$, $d = 9 - 4 = 5$
- Question 5 & 6, Exercise 4.6
- 1}{27}, \frac{1}{20}, \frac{1}{13}, \ldots \quad \text{ is in H.P.} \\ &27, 20, 13, \ldots \quad \text{ is in A.P.} \end{align*} Here $a_1 = 27$, $d = 20 - 27... rac{1}{3}, \frac{1}{3 \frac{1}{2}}, \ldots \quad \text{ is in H.P.} \\ &2, 2 \frac{1}{2}, 3, 3 \frac{1}{2}, \ldots \quad \text{ is in A.P.} \end{align*} Here $a_1 = 2$, $d = 2
- Question 1 and 2, Exercise 4.6
- $\frac{1}{9}, \frac{1}{12}, \frac{1}{15}, \cdots \text{ is in H.P.}$$ $$9, 12, 15, ... \text{ is in A.P.}$$ Here $a_1=9$, $d=12-9=3$, $a_7=?$. Gneral term of... . $$ Thus \begin{align*} a_7&=9+(6)(3) \\ & = 27 \text{ is in A.P} \end{align*} Hence the 7th term in H.... ac{1}{11}, \frac{1}{9}, \frac{1}{7}, \ldots\quad \text{ is in H.P.}\\ &11, 9, 7, \ldots \quad \text{ is
- Question 9 & 10, Exercise 4.6
- rac{1}{7}, \frac{1}{6}, -1, -\frac{1}{3}, \ldots \text{ is in H.P.}$$ $$7, 6, -1, -3, \ldots \text{ is in A.P.}$$ Here $a_1 = 7$, $d = 6 - 7 = -1$, $a_8=?$ T... n*} a_8 &= 7 + (7)(-1) \\ &= 7 - 7 \\ &= 0 \quad \text{is in A.P.} \end{align*} Hence, $a_8 = \dfrac{1}{... H = G^2$ is verified. GOOD ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:so
- Question 7 & 8, Exercise 4.6
- &= \frac{1}{4} - \frac{39}{28}\\ &= -\frac{8}{7} \text{ is in A.P.} \end{align*} Thus, the 14th term in ... {align*} a_{17}&=7+(16)(-3) \\ &= 7-48 \\ &= -41 \text{ is in A.P.} \end{align*} Hence 17th term in H.P is $-\dfrac{1}{41}$. GOOD ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-6-p3|< Question 5 & 6]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 11 and 12, Exercise 4.2
- of the saved money is $$1000, 3000, 5000, \dots, \text{ upto 20 terms}.$$ This is in A.P with $a_1 = 100... l be there in the 8th row. GOOD ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-2-p6|< Question 9 & 10]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit04:ex4-2-p8|Question 13 >]]</btn></text
- Question 14 and 15, Exercise 4.2
- ** Let $a= b$ and $b=20$. Then \begin{align*} &\text{A.M.} = \frac{a + b}{2} \\ \implies & 10 = \frac{... } Hence $x=-9$ and $y=24$. GOOD ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-2-p8|< Question 13]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit04:ex4-2-p10|Question 16 & 17 >]]</btn>
- Question 7 and 8, Exercise 4.3
- numbers from $2$ to $100$ is $$2+4+6+...+100 (50 \text{ terms}).$$ This is arithmetic series with: $a_{... $ to $100$ is $2550$. GOOD ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-3-p3|< Question 5 & 6]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit04:ex4-3-p5|Question 9 & 10 >]]</btn></
- Question 9 and 10, Exercise 4.3
- d numbers from $1$ to $99$ is $$1+3+5+...+99 (50 \text{ terms}).$$ This is arithmetic series with: $a_{... required sum is $34036$. GOOD ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-3-p4|< Question 7 & 8]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit04:ex4-3-p6|Question 11 & 12 >]]</btn><
- Question 26 and 27, Exercise 4.4
- $ each year. What will the population be in $15^{\text {th }}$ years? ** Solution. ** Here, \begin{al... ars will be $155,797$. GOOD ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-4-p12|< Question 24 & 25]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit04:ex4-4-p14|Question 28 & 29 >]]</btn><
- Question 11, Exercise 4.6
- and $b=\dfrac{4}{7}$, therefore \begin{align*} \text{H.M.}&=\frac{2ab}{a+b} \\ &=\frac{2\times\frac{2}... $ and $\dfrac{4}{7}$. GOOD ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-6-p5|< Question 9 & 10]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:unit04:ex4-6-p7|Question 12 >]]</btn></text>