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- Question 20 and 21, Exercise 4.4
- _\_ , \_\_\_ , \_\_\_ , 48$$ ** Solution. ** We have given $a_1=3$ and $a_5=48$. Assume $r$ be common... ference, then by general formula for nth term, we have $$ a_n=ar^{n-1}. $$ This gives \begin{align*} &a_... ="true"> **The good solution is as follows:** We have given $a_1=3$ and $a_5=48$. Assume $r$ be common... ference, then by general formula for nth term, we have $$ a_n=ar^{n-1}. $$ This gives \begin{align*} &a_
- Question 28 and 29, Exercise 4.4
- tance fallen. How far (up and down) will the ball have travelled when it hits the ground for the 6th tim... are in staff at high school? ** Solution. ** We have given \\ first person $=a_1= 1$ principal \\ Peop... \\ ... \\ People in 6ht round $= a_7$.\\ Thus we have the series $$ 1+2+4+...+a_7 $$ We have to find sum of geometric series with $a_1=1$, $r=2$, $n=7$. As
- Question 11 and 12, Exercise 4.8
- + Bk \ldots (2) \end{align*} Put $k=0$ in (2), we have \begin{align*} &1=2A + 0 \\ \implies & A = \frac{... \end{align*} Put $k+2=0 \implies k=-2$ in (2), we have \begin{align*} &1=0-2B\\ \implies &B = -\frac{1}{... {1}{k+2} \right). \end{align*} Taking the sum, we have \begin{align*} S_n &= \sum_{k=1}^n T_k = \frac{1}... Put $3k-2=0$ $\implies k=\dfrac{2}{3}$ in (2), we have \begin{align*} &1 = \left(3\times\frac{2}{3}+1 \r
- Question 1, Exercise 4.2
- ** Solution. ** Given: $a_1= 16$, $d=-2$.\\ We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&... ** Solution. ** Given: $a_1= 38$, $d=-4$.\\ We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&... Given: $a_1=\frac{3}{4}$, $d=\frac{1}{4}$.\\ We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&... Given: $a_1=\frac{3}{8}$, $d=\frac{5}{8}$.\\ We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2
- Question 9 and 10, Exercise 4.2
- cdots (ii) \end{align*} Comparing (i) and (ii) we have\\ \begin{align*} b-\frac{1}{a}&=\frac{1}{c}-b\\ b... \end{align*} Putting the value of $b$ in (i), we have\\ \begin{align*} d&=\frac{a+c}{2ac}-\frac{1}{a}\\... h second? ** Solution. ** By the given data, we have \begin{align*}16, 48, 80, \cdots \end{align*} This is an A.P with $a_1=16$ and $d=48-16=32$. We have $$a_n = a_1 + (n - 1) d.$$ This gives \begin{alig
- Question 23 and 24, Exercise 4.3
- ogether? ** Solution. ** From the statement, we have the following series: $$ 14+16+18+...+a_{25}.$$ T... tic series with $a_1=14$, $d=16-14=2$, $n=25$. We have to find $a_25$ and $S_25$.\\ As \begin{align} a_n... t layer? ** Solution. ** From the statement, we have the following series: $$ 50+49+48+...+6.$$ This i... c series with $a_1=50$, $d=49-50=-1$, $a_n=6$. We have to find $n$ and $S_n$.\\ As \begin{align} a_n&=a_
- Question 25 and 26, Exercise 4.3
- in all? ** Solution. ** From the statement, we have the following series: $$ 6000+70,000+...+a_{20}.$... $a_1=6,000$, $d=70,000-6,000=64,000$, $n=20$. We have to find $S_n$.\\ As \begin{align} S_n&=\frac{n}{2... 31 days) ** Solution. ** From the statement, we have the following series: $$ 500+550+600+...+a_{31}.$... series with $a_1=500$, $d=550-500=50$, $n=31$. We have to find $S_n$.\\ As \begin{align} S_n&=\frac{n}{2
- Question 22 and 23, Exercise 4.4
- \_\_, \_\_\_, \dfrac{1}{4}$$ ** Solution. ** We have $a_1=8$ and $a_6=\frac{1}{4}$. Assume $r$ to be ... en, by the general formula for the $n$th term, we have \\ $a_n = a_1 r^{n-1}.$\\ This gives\\ \begin{ali... means. $$3 , \_\_\_ , 75$$ ** Solution. ** We have $a_1=3$ and $a_3=75$. Assume $r$ to be the comm... en, by the general formula for the $n$th term, we have $a_n = a_1 r^{n-1}.$ This gives \begin{align*} a
- Question 24 and 25, Exercise 4.4
- \_\_\_, \_\_\_, \_\_\_, 80$$ ** Solution. ** We have $a_1=5$ and $a_5=80$. Assume $r$ to be the comm... en, by the general formula for the $n$th term, we have \\ $$a_n = a_1 r^{n-1}.$$ This gives \begin{align... _\_\_, \_\_\_, \_\_\_, 112$$ ** Solution. ** We have $a_1=7$ and $a_6=112$.\\ Assume $r$ to be the c... en, by the general formula for the $n$th term, we have \\ $$a_n = a_1 r^{n-1}.$$ This gives \begin{alig
- 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}+1 \r... ut $3k+1=0$ $\implies k=-\dfrac{1}{3}$ in (2), we have \begin{align*} &1 = 0+\left(3\left(-\frac{1}{3}\r... Put $5k-4=0$ $\implies k=\dfrac{4}{5}$ in (2), we have \begin{align*} &1 = \left(5\times\frac{4}{5}+1 \r... ut $5k+1=0$ $\implies k=-\dfrac{1}{5}$ in (2), we have \begin{align*} &1 = 0+\left(5\left(-\frac{1}{5}\r
- Question 2, Exercise 4.2
- 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_5 &=5... $, $d=14-11=3$.\\ Now $$a_n=a_1+(n-1)d.$$ So, we have \begin{align*} a_4 &= 11 + (4-1) \cdot 3 = 11 + 9... = 4\). Now $$a_n = a_1 + (n - 1)d.$$ So, we have \[ \begin{aligned} a_4 &= -5.4 + (4 - 1)(4) = -
- Question 17, 18 and 19, Exercise 4.3
- o, $a_{1}=6$, $d=12-6=6$, $a_{n}=96$, $n=?$.\\ We have \begin{align} & a_n=a_1+(n-1)d \\ \implies & 96... o, $a_{1}=34$, $d=30-34=-4$, $a_{n}=2$, $n=?$. We have \begin{align} & a_n=a_1+(n-1)d \\ \implies & 2=... , $a_{1}=10$, $d=4-10=-6$, $a_{n}=-50$, $n=?$. We have \begin{align} & a_n=a_1+(n-1)d \\ \implies & -5
- Question 20, 21 and 22, Exercise 4.3
- lies & n=\frac{1752}{146}=12. \end{align} Also we have \begin{align} &a_n=a_1+(n-1)d\\ \implies & 139=7+... _1=378-371 \\ \implies a_1=1. \end{align} Also we have \begin{align} &a_n=a_1+(n-1)d\\ \implies & 53=1+(... lies & n=\frac{3432}{312}=11. \end{align} Also we have \begin{align} &a_n=a_1+(n-1)d\\ \implies & 306=6+
- Question 21 and 22, Exercise 4.7
- 1$. \\ Also kth term of $1+2+3+...$ is $k$. So we have required kth term $k(3k+1)$. </callout> Conside... ) \\ &=3k^2+k. \end{align*} Taking summation, we have \begin{align*}\sum_{k=1}^{n} T_{k} &= \sum_{k=1}^... 8k^3+12k^2-2k-3\end{align*} Taking summation, we have \begin{align*}\sum_{k=1}^{n} T_{k} &= \sum_{k=1}^
- Question 11 and 12, Exercise 4.2
- A.P with $a_1=35$, $d=31-35=-4$ and $a_8=?$.\\ We have $$a_n=a_1+(n-1)d.$$ Thus, we have \begin{align*} a_8 &= 35 + (7)(-4) \\ &=35-28\\ &=7. \end{align*} Hen