Search
You can find the results of your search below.
Fulltext results:
- 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
- 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 2 Exercise 4.5
- a_n, n_2 r$ and $S_n$ of a geometric sequence are given. Find the ones that are missing $a_1=1, \quad r=-... a_n, n_2 r$ and $S_n$ of a geometric sequence are given. Find the ones that are missing $r=\dfrac{1}{2}, ... {\prime \prime}]}{1-r},\end{align} becomes in the given case\\ \begin{align}\Rightarrow S_9&=\dfrac{2^8[1... a_n, n_2 r$ and $S_n$ of a geometric sequence are given. Find the ones that are missing $r=-2, S_n=-63, a
- Question 3 and 4 Exercise 4.1
- 4}{5}, \ldots$ ====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 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 Exercise 4.4
- te the first five terms of geometric ric sequence given that $a_1=5, \quad r=3$ ====Solution==== The gcom... te the first five terms of geometric ric sequence given that $a_1=8, \quad r=-\dfrac{1}{2}$ ====Solution=... te the first five terms of geometric ric sequence given that $a_1=-\dfrac{9}{16}, \quad r=-\dfrac{2}{3}$ ... te the first five terms of geometric ric sequence given that $a_1=\dfrac{x}{y}, \quad r=-\dfrac{y}{x}$ ==
- Question 1 Exercise 4.5
- sum $3+6+12+\ldots+3.2^9$ ====Solution==== In the given geometric series: $a_1=3, \quad r=\dfrac{6}{3}=2$... dfrac{a_1(r^n-1)}{r-1},\end{align} becomes in the given case\\ \begin{align}S_{10}&=\dfrac{3[2^{10}-1]}{2... r^n-1)}{r-1} \text {, }\end{align} becomes in the given case\\ \begin{align}S_8&=\dfrac{8[1-(\dfrac{1}{2}... $$S_7=\dfrac{a_1(r^n-1)}{r-1}$$,\\ becomes in the given case\\ \begin{align}\Rightarrow S_7&=\dfrac{2^4[2
- Question 4 Exercise 4.5
- quad r=0.01<1$.\\ Therefore the sum exists and is given by\\ \begin{align}S_{\infty}&=\dfrac{a_1}{1-r}\\ ... with $a_1=0.15, r=0.01<1$.\\ Thus the sum of the given series exists and given by $$S_{\infty}=\dfrac{a_1}{1-r}$$,\\ putting $a_1, r$ we get\\ $$S_{\infty}=\df... r=0.001<1 \end{align} Thus the sum exists and is given by\\ \begin{align}S_{\infty}&=\dfrac{a_1}{1-r}\en
- Question 5 and 6 Exercise 4.2
- efore $$a_n=\log (a b^{n-1}).$$ We show that the given sequence is A.P. Since \begin{align}a_n&=\log(a ... is constant, i.e. independent of $n$. Thus, the given sequence is in A.P. GOOD =====Question 6===== Fi... ind the sequence. GOOD ====Solution==== Since the given terms are in A.P, \begin{align}& (6 k-2)-(2 k+7)=
- Question 1 and 2 Exercise 4.2
- &=2+42=44 \end{align} Hence the 15th term of the given sequence is $44$. =====Question 2===== The first... _7=8+30=38 .\end{align} Hence the 7th term of the given sequence is 38. ====Go To==== <text align="righ
- Question 7 Exercise 4.2
- irst term and $d$ be common difference of A.P. As given \begin{align} &a_6+a_4=6 \\ \implies & a_1+5d+a_... lies & a_1+4d=3 --- (1) \end{align} Also, we have given \begin{align} &a_6-a_4=\dfrac{2}{3} \\ \implies &
- Question 15 Exercise 4.2
- en $$ A=\dfrac{a+b}{2}. --- (1) $$ Also, we have given $$ A=\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}. --- (2) $$... align} Hence $n=0$, when $a\neq b$ and for $a=b$, given expression is A.M for all $n$. ====Go To==== <tex
- Question 1 Exercise 4.3
- _1$ be first term and $d$ be common difference of given A.P. Then \begin{align}&a_1=9 \\ &d=7-9=-2 \\ &n... _1$ be first term and $d$ be common difference of given A.P. Then \begin{align}&a_1=3 \\ &d=\dfrac{8}{3}
- Question 7 & 8 Exercise 4.3
- o $3 n$ terms. ====Solution==== We can reform the given series into three arithmetic series \begin{align}... ch one will give us sum of the $3 n$ terms of the given series.\\ For $$1+7+13+\ldots$$ here $$a_1=1, d=7
- Question 5 & 6 Exercise 4.5
- 1}$$\\ Putting both the $S_{10}$ and $S_S$ in the given equation, we get\\ \begin{align}\dfrac{a_1(r^{10}... lign}\\ Puting (i),(ii) and (iii) in L.H.S of the given, we get \begin{align}S_n(S_{3 n}-S_{2 n})&=\dfrac