Search
You can find the results of your search below.
Fulltext results:
- Question 11 and 12, Exercise 4.8
- \end{align*} Multiplying both sides by $k(k+2)$, we get \begin{align*} 1 = A(k+2) + Bk \ldots (2) \end{align*} Put $k=0$ in (2), we have \begin{align*} &1=2A + 0 \\ \implies & A = \... }. \end{align*} Put $k+2=0 \implies k=-2$ in (2), we have \begin{align*} &1=0-2B\\ \implies &B = -\fra... Using the values of $A$ and $B$ in equation (1), we get \begin{align*} \frac{1}{k(k+2)} &= \frac{1}{2
- Question 20 and 21, Exercise 4.4
- \_\_\_ , \_\_\_ , \_\_\_ , 48$$ ** Solution. ** We have given $a_1=3$ and $a_5=48$. Assume $r$ be c... difference, then by general formula for nth term, we have $$ a_n=ar^{n-1}. $$ This gives \begin{align*... con="true"> **The good solution is as follows:** We have given $a_1=3$ and $a_5=48$. Assume $r$ be c... difference, then by general formula for nth term, we have $$ a_n=ar^{n-1}. $$ This gives \begin{align*
- Question 14 and 15, Exercise 4.2
- ====Question 14===== Find '$b$' if $10$ is A.M between $b$ and $20$. ** Solution. ** Let $a= b$ and... callout type="warning" icon="true"> To get the answer given at the end of the book, question can be as follows: * Find '$b$' if $25$ is A.M between $b$ and $20$. * Find '$b$' if $10$ is A.M between $b$ and $-10$. </callout> =====Question 15====
- Question 20, 21 and 22, Exercise 4.3
- Given $a_{1}=7$, $a_{n}=139$, $S_{n}=876$. First we find $n$ and $d$.\\ As \begin{align} &S_n=\frac{n... implies & n=\frac{1752}{146}=12. \end{align} Also we have \begin{align} &a_n=a_1+(n-1)d\\ \implies & 1... ** Given $n=14$, $a_{n}=53$, $S_{n}=378$. First we find $a_1$ and $d$.\\ As \begin{align} &S_n=\frac... 7a_1=378-371 \\ \implies a_1=1. \end{align} Also we have \begin{align} &a_n=a_1+(n-1)d\\ \implies & 5
- Question 22 and 23, Exercise 4.4
- \_\_\_, \_\_\_, \dfrac{1}{4}$$ ** Solution. ** We have $a_1=8$ and $a_6=\frac{1}{4}$. Assume $r$ t... Then, by the general formula for the $n$th term, we have \\ $a_n = a_1 r^{n-1}.$\\ This gives\\ \begi... \\ \implies r &= \frac{1}{2}. \end{align*} Thus, we can find the missing terms: \begin{align*} a_2 &=... ric means. $$3 , \_\_\_ , 75$$ ** Solution. ** We have $a_1=3$ and $a_3=75$. Assume $r$ to be the
- Question 24 and 25, Exercise 4.4
- , \_\_\_, \_\_\_, \_\_\_, 80$$ ** Solution. ** We have $a_1=5$ and $a_5=80$. Assume $r$ to be the... Then, by the general formula for the $n$th term, we have \\ $$a_n = a_1 r^{n-1}.$$ This gives \begin{... r^4 &= 16 \\ \implies r &= 2. \end{align*} Thus, we can find the missing terms: \begin{align*} a_2 &=... ,\_\_\_, \_\_\_, \_\_\_, 112$$ ** Solution. ** We have $a_1=7$ and $a_6=112$.\\ Assume $r$ to be
- Question 14, Exercise 4.5
- te geometric series; $0.444...$ ** Solution. ** We can express the decimal as $$0.444... = 0.4+0.04+... geometric series; $9.99999 ...$ ** Solution. ** We can express the decimal as $$0.99999 ... = 0.9+0.... ometric series; $0.5555 \ldots$ ** Solution. ** We can express the decimal as $$0.5555 \ldots = 0.5 ... ometric series; $0.6666 \ldots$ ** Solution. ** We can express the decimal as $$0.6666 \ldots = 0.6
- Question 7 and 8, Exercise 4.8
- *} Put $3k-2=0$ $\implies k=\dfrac{2}{3}$ in (2), we have \begin{align*} &1 = \left(3\times\frac{2}{3}... w put $3k+1=0$ $\implies k=-\dfrac{1}{3}$ in (2), we have \begin{align*} &1 = 0+\left(3\left(-\frac{1}... \end{align*} Using value of $A$ and $B$ in (1), we get \begin{align*} \frac{1}{(3k-2)(3k+1)} = \frac... *} Put $5k-4=0$ $\implies k=\dfrac{4}{5}$ in (2), we have \begin{align*} &1 = \left(5\times\frac{4}{5}
- Question 28 and 29, Exercise 4.4
- 8===== A super ball dropped from the top of the tower (556 ft high) always rebounds three-fourths of t... ple are in staff at high school? ** Solution. ** We have given \\ first person $=a_1= 1$ principal \\... =4$ \\ ... \\ People in 6ht round $= a_7$.\\ Thus we have the series $$ 1+2+4+...+a_7 $$ We have to find sum of geometric series with $a_1=1$, $r=2$, $n=7
- Question 1, Exercise 4.2
- 2$. ** Solution. ** Given: $a_1= 16$, $d=-2$.\\ We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*}... 4$. ** Solution. ** Given: $a_1= 38$, $d=-4$.\\ We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*}... ** Given: $a_1=\frac{3}{4}$, $d=\frac{1}{4}$.\\ We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*}... ** Given: $a_1=\frac{3}{8}$, $d=\frac{5}{8}$.\\ We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*}
- Question 2, Exercise 4.2
- a_1=5$, $d=9-5=4$.\\ Now $$a_n=a_1+(n-1)d.$$ So, we have \begin{align*} a_4 &=5+(4-1)(4)=5+12=17\\ a... =11$, $d=14-11=3$.\\ Now $$a_n=a_1+(n-1)d.$$ So, we have \begin{align*} a_4 &= 11 + (4-1) \cdot 3 = 1... {1}{2}, \frac{3}{2}, \frac{5}{2}, \ldots$$ First, we find the common difference: \begin{align*} d &= \... 4) = 4\). Now $$a_n = a_1 + (n - 1)d.$$ So, we have \[ \begin{aligned} a_4 &= -5.4 + (4 - 1)(4
- Question 9 and 10, Exercise 4.2
- b \cdots (ii) \end{align*} Comparing (i) and (ii) we have\\ \begin{align*} b-\frac{1}{a}&=\frac{1}{c}-... ac} \end{align*} Putting the value of $b$ in (i), we have\\ \begin{align*} d&=\frac{a+c}{2ac}-\frac{1}... 8 th second? ** Solution. ** By the given data, we have \begin{align*}16, 48, 80, \cdots \end{align*} This is an A.P with $a_1=16$ and $d=48-16=32$. We have $$a_n = a_1 + (n - 1) d.$$ This gives \begin
- Question 13, Exercise 4.2
- akistan. =====Question 13(i)===== Find A.M. between $7$ and $17$ ** Solution. ** Here $a=7$ and $... 12$. GOOD =====Question 13(ii)===== Find A.M. between $3+3 \sqrt{2}$ and $7-3 \sqrt{2}$ ** Solution.... 5$. GOOD =====Question 13(iii)===== Find A.M. between $7 \sqrt{5}$ and $\sqrt{5}$ ** Solution. ** He... 5}$. GOOD =====Question 13(iv)===== Find A.M. between $2y+5$ and $5y+3$ ** Solution. ** Here $a=2y+5
- Question 16 and 17, Exercise 4.2
- Question 16===== Find the two arithmetic means between $5$ and $17$. ** Solution. ** Let $A_1$ and $A_2$ be two arithmetic means between $5$ and $17$.\\ Then $5$, $A_1$, $A_2$, $17$ ar... Question 16===== Find the two arithmetic means between $5$ and $17$. ** Solution. ** Let $A_1$, $A_2$ and $A_3$ be thre arithmetic means between $2$ and $-18$.\\ Then $2$, $A_1$, $A_2$, $A_3$,
- Question 23 and 24, Exercise 4.3
- altogether? ** Solution. ** From the statement, we have the following series: $$ 14+16+18+...+a_{25}... hmetic series with $a_1=14$, $d=16-14=2$, $n=25$. We have to find $a_25$ and $S_25$.\\ As \begin{align... last layer? ** Solution. ** From the statement, we have the following series: $$ 50+49+48+...+6.$$ T... etic series with $a_1=50$, $d=49-50=-1$, $a_n=6$. We have to find $n$ and $S_n$.\\ As \begin{align} a_