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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 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\\ $$1.63=1+\dfrac{7}{11}=\dfrac{18}{11} \text
- Question 6 & 7 Exercise 4.4
- 6}=n;\quad$ show that $\ln =m^2$ ====Solution==== We know that $a_n=a_1 r^{n-1}$ therefore\\ \begin{al... hbf{A 5}}=n\end{align} Multiplying (i) and (iii), we get\\ \begin{align}a_{10} \cdot a_{16}&=\ln =(a_1... o form a geometric sequence. ====Solution==== Let we are considering the standard geometric sequence w... s, a_1 r^{n-1},$$ Taking reciprocal of the terms, we get $\dfrac{1}{a_1}, \dfrac{1}{a_1 r}, \dfrac{1}{
- Question 1 Exercise 4.5
- a_1=3, \quad r=\dfrac{6}{3}=2$ 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 17 Exercise 4.2
- 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... mplies n&=\dfrac{27}{d}-1 ---(i)\end{align} Also, we have given $A_3:A_7=7:13$, where $$A_3=a_4=a_1+3d=5+3d$$ and $$A_7=a_8=a_1+7d=5+7d.$$ Thus we have \begin{align}&\dfrac{A_3}{A_7}=\dfrac{7}{13}
- Question 2 & 3 Exercise 4.4
- Here $$a_3=27 \quad\text{and}\quad a_5=243$$ and we know\\ \begin{align}a_3&=a_1 r^2=27\\ a_5&=a_1 r^4=243.\end{align} Dividing (ii) by (i), we get\\ \begin{align}\dfrac{a_1 r^4}{a_1 r^2}&=\dfr... -\sqrt{2}...(ii)\end{align} Dividing (ii) by (i), we get\\ \begin{align}\dfrac{a_1 r^2}{a_1 r}&=-\dfra... =2 \\ \Rightarrow a_1&=-2 \sqrt{2}\end{align} Now we know that\\ \begin{align}a_n&=a_1 r^{n-1}, \\ \te
- 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 ... 1 and so on, therefore $$a_n=\log (a b^{n-1}).$$ We show that the given sequence is A.P. Since \begin{align}a_n&=\log(a b^{n-1}). \end{align} Now we take \begin{align} d&=a_{n+1}-a_n \\ &=\log (a b^... {a b^n}{a b^{n-1}}\right)\\ &=\log b. \end{align} We see that the difference of consecutive terms $d$
- Question 15 Exercise 4.2
- $b$, 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}+... & b^n(a-b)=a^n(a-b)\end{align} If $a\neq b$, then we have \begin{align} &b^n =a^n \\ \implies &\dfr... implies &n=0.\end{align} If $a=b$, then from (3), we have \begin{align}&\dfrac{a+a}{2}=\dfrac{a^{n+1}+
- Question 7 & 8 Exercise 4.3
- -$ $17+\ldots$ upto $3 n$ terms. ====Solution==== We can reform the given series into three arithmetic... e series each one in parenthesis is arithmetic,\\ we can find the sum of $\mathrm{n}$ terms of each, a... . }\end{align} Putting (i),(ii) and (iii) in (1), we get\\ \begin{align}S_{3 n}&=S_n+S_n^{\prime}-(S_n... term $a_1=1$, and the common difference $d=2$.\\ We know that: \begin{align}S_n&=\dfrac{n}{2}[2 a_1+(
- Question 1 Exercise 4.4
- a_1 r^3, a_1 r^4, \ldots$, so for $a_1=5 ; r=3$, we have \begin{align}&5,5.3,5.3^2, 5.3^3, 5.3^4, \ld... ^4, \ldots$,\\ so for $$a_1=8 ; r=-\dfrac{1}{2}$$ we have\\ \begin{align}&8,8(-\dfrac{1}{2}), 8(-\dfra... so for $$a_1=-\dfrac{9}{16} ; r=-\dfrac{2}{3}$$ we have\\ \begin{align}&-\dfrac{9}{16} ,-\dfrac{9}{1... s$,\\ so for $$a_1=\dfrac{x}{y} r=-\dfrac{y}{x}$$ we have,\\ \begin{align}&\dfrac{x}{y}, \dfrac{x}{y}
- Question 3 Exercise 4.5
- nce having $a_2=2$ and $a_3=1$\\ ====Solution==== We first try to have find $a_1$ and $r$.\\ We know that $$a_n=a_1 r^{n-1}$$\\ therefore\\ $$a_2=a_1 r=2.... $$a_3=a_1 r^2=1...(ii)$$\\ Dividing (ii) by (i), we get\\ \begin{align}\dfrac{a_1 r^2}{a_1 r}&=\dfrac... {1}{2} \text {, }\end{align} putting this in (i), we have\\ \begin{align}\dfrac{a_1}{2}&=2\\ \Rightarr
- Question 13 & 14 Exercise 4.5
- ===== Adding 1 to both sides of the given series, we get $$1+y=1+\dfrac{x}{3}+\dfrac{x^2}{3^2}+\dfrac{... infty}=\dfrac{a_1}{1-r}$, putting $a_1, \quad r$, we get $$S_{\infty}=\dfrac{1}{1-\dfrac{x}{3}}=\dfrac{3}{3-x}$$ putting in (i), we have \begin{align}1+y&=\dfrac{3}{3-x} \\ \Rightar... 1875\\ &=9.921875\end{align} Putting (ii) in (i), we get\\ \begin{align}S&=10+2(9.921875) \\ & =10+19.
- Question 3 and 4 Exercise 4.1
- rac{3}{4}, \dfrac{4}{5}, \ldots$ ====Solution==== We can reform the given sequence to pick the pattern... pattern. $2,-4,6,-8,10, \ldots$ ====Solution==== We can reform the given sequence to pick the pattern... the pattern. $1,-1,1,-1, \ldots$ ====Solution==== We can reform the give sequence to pick the pattern