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- Question 3 & 4 Exercise 4.3
- t \begin{align}a_n&=a_1+(n-1) d\end{align} in the given case it becomes,\\ \begin{align} 350&=25+(n-1)(5)
- Question 2 Exercise 4.3
- of the components $a_1, a_n, n, d$ and $S_n$ are given. Find the one that is missing: $a_1=2, n=17, d=3$. GOOD ====Solution==== Given: $a_1=2, n=17, d=3$ \\ We need to find $a_{17}$ a... of the components $a_1, a_n, n, d$ and $S_n$ are given. Find the one that are missing $a_1=-40, S_{21}=210$. GOOD ====Solution==== Given: $a_1=-40$ and $S_{21}=210$.\\ So we have $n=21$
- Question 1 Exercise 4.3
- _1$ be first term and $d$ be common difference of given A.P. Then \begin{align}&a_1=9 \\ &d=7-9=-2 \\ &n... _1$ be first term and $d$ be common difference of given A.P. Then \begin{align}&a_1=3 \\ &d=\dfrac{8}{3}
- Question 17 Exercise 4.2
- &=\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
- Question 15 Exercise 4.2
- en $$ A=\dfrac{a+b}{2}. --- (1) $$ Also, we have given $$ A=\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}. --- (2) $$... align} Hence $n=0$, when $a\neq b$ and for $a=b$, given expression is A.M for all $n$. ====Go To==== <tex
- Question 12 & 13 Exercise 4.2
- .$$ Increase in salary in each year $=d=750$. The given problem is of A.P and we have to find $a_{21}$.
- Question 11 Exercise 4.2
- climb in each succeeding hour $= d=100$. As the given problem is of A.P with $a_n=5400$, we have to fin
- Question 10 Exercise 4.2
- $ Decrease in population each year=$d=-500$. The given problem is of A.P and we have to find population
- Question 9 Exercise 4.2
- a_2=21,$$ $$a_3=18.$$ Since $$d=21-24=18-21=-3,$$ given sequence is in A.P. Now \begin{align} a_8&=a_1+7d
- Question 7 Exercise 4.2
- irst term and $d$ be common difference of A.P. As given \begin{align} &a_6+a_4=6 \\ \implies & a_1+5d+a_... lies & a_1+4d=3 --- (1) \end{align} Also, we have given \begin{align} &a_6-a_4=\dfrac{2}{3} \\ \implies &
- Question 5 and 6 Exercise 4.2
- efore $$a_n=\log (a b^{n-1}).$$ We show that the given sequence is A.P. Since \begin{align}a_n&=\log(a ... is constant, i.e. independent of $n$. Thus, the given sequence is in A.P. GOOD =====Question 6===== Fi... ind the sequence. GOOD ====Solution==== Since the given terms are in A.P, \begin{align}& (6 k-2)-(2 k+7)=
- Question 3 and 4 Exercise 4.2
- =24+1=25.\end{align} Thus, the number of terms in given progression are $25$. GOOD =====Question 4===== The $n$th term of sequence is given by $a_n=2n+7$. Show that it is an arithmetic prog... ression. Also find its 7th term. ====Solution==== Given that $$a_n=2 n+7. --- (1)$$ Then \begin{align}a_{... get $$a_7=2(7)+7=14+7=21.$$ Hence the 7th term of given AP is 21. GOOD ====Go To==== <text align="left"><
- Question 1 and 2 Exercise 4.2
- &=2+42=44 \end{align} Hence the 15th term of the given sequence is $44$. =====Question 2===== The first... _7=8+30=38 .\end{align} Hence the 7th term of the given sequence is 38. ====Go To==== <text align="righ
- Question 3 and 4 Exercise 4.1
- 4}{5}, \ldots$ ====Solution==== We can reform the given sequence to pick the pattern of the sequence as: ... -8,10, \ldots$ ====Solution==== We can reform the given sequence to pick the pattern of the sequence as: ... ively. $a_1=3$, $a_{n+1}=5-a_n$. ====Solution==== Given $$a_1=3, a_{n+1}=5-a_n.$$ For $n=1$ \begin{align}... $a_1=3, a_{n+1}=\dfrac{a_n}{n}$ ====Solution==== Given $$a_1=3, a_{n+1}=\frac{a_n}{n}$$ For $n=1$ \begin
- Question 1 and 2 Exercise 4.1
- == Find first four terms of the sequence with the given general terms: $a_n=\dfrac{n(n+1)}{2}$ ====Solution==== Given: $$a_n=\dfrac{n(n+1)}{2}$$ For first term, put $n... == Find first four terms of the sequence with the given general terms: $a_n=(-1)^{n-1} 2^{n+1}$ ====Solution==== Given: $$a_n=(-1)^{n-1} 2^{n+1}$$ For first term, put