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- Question 2 Exercise 4.3
- OOD ====Solution==== Given: $a_1=2, n=17, d=3$ \\ We need to find $a_{17}$ and $S_{17}$. As we know $$a_{n}=a_1+(n-1)d.$$ Thus $$a_{17}=2+(17-1)(3)=50.$$ ... ution==== Given: $a_1=-40$ and $S_{21}=210$.\\ So we have $n=21$ and we have to find $a_{21}$ and $d$. As \begin{align}&S_{21}=\dfrac{21}{2}(a_1+a_{21}) \
- Question 16 Exercise 4.2
- =Question 16===== Insert five arithmetic means between $5$ and $8$ and show that their sum is five times the arithmetic mean between $5$ and $8$. GOOD ====Solution==== Let $A_1, A_2, A_3, A_4, A_5$ be five arithmetic means between $5$ and $8$. Then $5, A_1, A_2, A_3, A_4, A_5, ... re in A.P, where $$a_1=5 \text{ and } a_7=8.$$ As we have \begin{align}&a_7=a+6d\\ \implies &8=5+6d\\
- Question 12 & 13 Exercise 4.2
- ear for 20 years, what was his salary during his twenty first year of work? GOOD ====Solution==== Supp... ch year $=d=750$. The given problem is of A.P and we have to find $a_{21}$. As, we have \begin{align} a_{21}&=a_1+20d\\ &=3500+20(750) \\ &=18500. \end{a... ==Question 13(i)===== Find the arithmetic mean between $12$ and $18$. GOOD ====Solution==== Here $a=12
- Question 14 Exercise 4.2
- stion 14(i)===== Insert three arithmetic means between 6 and 41. GOOD ====Solution==== Let $A_1, A_2, A_3$ be three arithmetic means between 6 and 41. Then $6, A_1, A_2, A_3, 41$ are in A.P. We have $$a_1=6 \text{ and } a_6=41.$$ Now \begin{al... 1}{4}.\end{align} Hence three arithmetic means between 6 and 41 are $$14\dfrac{3}{4},23\dfrac{1}{2},32
- Question 17 Exercise 4.2
- uestion 17===== There are $n$ arithmetic means between 5 and 32 such that the ratio of the 3rd and 7th... A_2, A_3, \ldots, A_n$ be $n$ arithmetic means between 5 and 32. Then $5, A_1, A_2, A_3, \ldots, A_n, ... in A.P, where $$a_1=5 \text{ and } a_{n+2}=32.$$ We know $a_n=a_1+(n-1) d$, replacing $n$ by $n+2$, we get\\ \begin{align}a_{n+2}&=a_1+(n+2-1) d \\ & =a
- Question 6 & 7 Exercise 4.4
- 6}=n;\quad$ show that $\ln =m^2$ ====Solution==== We know that $a_n=a_1 r^{n-1}$ therefore\\ \begin{al... hbf{A 5}}=n\end{align} Multiplying (i) and (iii), we get\\ \begin{align}a_{10} \cdot a_{16}&=\ln =(a_1... =a_1 r^{12} \text { by (ii) }\end{align} Hence showed that $\ln =m^2$. =====Question 7===== Show that... o form a geometric sequence. ====Solution==== Let we are considering the standard geometric sequence w
- Question 10 Exercise 4.4
- ion 10===== Find two numbers if the difference between them is $48$ and their A.M exceeds their G.M by... $a$ and $b$ \\ Condition-$1$\\ The difference between them is $48$\\ Therefore, $$\quad a-b=48....(i)$$. The geometric mean between $a$ and $b$ is $$G=\sqrt{a b}$$ The arithmetic mean between $a$ and $b$ is $$A=\dfrac{a+b}{2}$$ Condition-$
- Question 4 Exercise 4.5
- mmon fraction $0 . \overline{8}$ ====Solution==== We can write $$0 . \overline{8}=0.888888 \ldots$$\\ ... geometric series with $$a_1=0.8, \quad r=0.1$$\\ We can find the infinite sum as:\\ $$S_{\infty}=\dfr... frac{8}{9}$$\\ Hence putting $S_{\infty}$ in (i), we get $$0 . \overline{8}=\dfrac{8}{9}$$.\\ =====Qu... . \text { (ii) }\end{align} Putting (ii) in (i), we get\\ $$1.63=1+\dfrac{7}{11}=\dfrac{18}{11} \text
- Question 5 and 6 Exercise 4.2
- an A.P. Also find its nth term. ====Solution==== We first find $n$th term. Each term of the sequence ... $ of some number. Each log contains $a$ but the power of $b$ in first term is zero, in second term the power of $b$ is 1 and so on, therefore $$a_n=\log (a b^{n-1}).$$ We show that the given sequence is A.P. Since \begi
- Question 15 Exercise 4.2
- n+1}+b^{n+1}}{a^n+b^n}$ is the arithmetic mean between $a$ and $b$. Where $a$ and $b$ are not zero sim... === Suppose $A$ represents the arithmetic mean between $a$ and $b$, then $$ A=\dfrac{a+b}{2}. --- (1) $$ Also, we have given $$ A=\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}. --- (2) $$ Comparing (1) and (2), we have \begin{align}&\dfrac{a+b}{2}=\dfrac{a^{n+1}+
- Question 13 & 14 Exercise 4.3
- ce $20,23,26, \ldots$ is an arithmetic equence.\\ We have to find the total number of seats that are $S_{40}$.\\ We know by sum formula\\ $$S_n=\dfrac{n}{2} [{2} a_1... uestion 14===== Insert enough arithmetic means between $1$ and $50$ so that the sum of the resulting series will be $459$ . ====Solution==== We know that:\\ \begin{align}S_n&=\dfrac{n}{2}[a_1+a
- Question 1 Exercise 4.5
- a_1=3, \quad r=\dfrac{6}{3}=2$ and $a_n=3.2^9$.\\ We first find $n$ and then the sum of series.\\ We know that $$a_n=a_1 r^{n-1}$$,\\ \begin{align}3.2^9&=... frac{1}{2} \quad$$\\ and $$a_n=\dfrac{1}{16}$$.\\ We first find $n$ and then the sum of series. We know that\\ \begin{align}a_n&=a_1 r^{n-1} \text {, }\\
- Question 2 Exercise 4.5
- _1=1, \quad r=-2, \quad a_n=64$. ====Solution==== We first find $n$ and then $S_n$\\ We know $a_n=a_1 r^{n-1}$, therefore\\ \begin{align}64&=(-2)^{n-1}\\ ... ing $r=\dfrac{1}{2}, a_9=1, n=9$ ====Solution==== We first find $a_1$ and then $S_9$.\\ We know $$a_9=a_1 r^8$$\\ therefore we have\\ \begin{align}1&=a_1
- Question 5 & 6 Exercise 4.5
- $r$ such that $S_{10}=244 S_5$. ====Solution==== We know that $$S_n=\dfrac{a_1(r^n-1)}{r-1}$$\\ then\... oth the $S_{10}$ and $S_S$ in the given equation, we get\\ \begin{align}\dfrac{a_1(r^{10}-1)}{r-1}&=24... _{3 n}-S_{2 n})=(S_n-S_{2 n})^2$ ====Solution==== We know that:\\ $$S_n=\dfrac{a_1(r^n-1)}{r-1}....(i)... lacing $n$ by $2 n$, and $3 n$ in the above, then we get\\ \begin{align}S_{2 n}&=\dfrac{a_1(r^{2 n}-1)
- Question 9 & 10 Exercise 4.3
- ion 9===== Find the sum 'of all multiples of 9 between 300 and 700. ====Solution==== All the multiples of 9 between 300 and 700 are:\\ $$306,315,324,333, \ldots, 6... lign} Hence, sum of all multiples of $9$ lying between $300$ and $700$ is equal to $21,978$. =====Que... l money for distribution $S_4=1000$, \\ therefore we have $n=4$\\ \begin{align}\text{Let the first per