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- Question 12 & 13 Exercise 4.2
- ====Solution==== Here $a=12, b=18$.\\ Let say $A$ be arithmetic means. Then \\ \begin{align}A&=\dfrac{... Here $a=\dfrac{1}{3}, b=\dfrac{1}{4}$,\\ Let $A$ be arithmetic mean. Then\\ \begin{align}A&=\dfrac{a+... ====Solution==== Here $a=-6, b=-216$.\\ Let $A$ be arithmetic mean. Then\\ $$A=\dfrac{a+b}{2}=\dfrac... \prime}=(a+b)^2$, $b^{\prime}=(a-b)^2$.\\ Let $A$ be arithmetic mean. Then\\ \begin{align} A&=\dfrac{a
- Question 1 Exercise 4.3
- h term; 20 terms. GOOD ====Solution==== Let $a_1$ be first term and $d$ be common difference of given A.P. Then \begin{align}&a_1=9 \\ &d=7-9=-2 \\ &n=20... h term; 11 terms. GOOD ====Solution==== Let $a_1$ be first term and $d$ be common difference of given A.P. Then \begin{align}&a_1=3 \\ &d=\dfrac{8}{3}-3=
- Question 15 & 16 Exercise 4.5
- amount each day. If this continụed, how much must be set aside on the $15^{\text {th }}$ day? What is ... daily rate of increase if the rate is assumed to be constant. ====Solution==== The number of bacteria... _1=64000$ Let number of bacteria after first day be $a_2$, after second be $a_3$ and so on. The number of bacteria after $6$th day $a_7=729000$. Since t
- Question 9 & 10 Exercise 4.5
- r&=1 \text { or } r=2.\end{align} But $r$ can not be 1 thus $r=3$. =====Question 10==== How many terms of the series: $1+\sqrt{3}+3+\ldots$ be added to get the sum $40+13 \sqrt{3}$ ====Solutio... of the given sequence are added then the sum will be $40+13 \sqrt{3}$. ====Go To==== <text a
- Question 7 Exercise 4.2
- thmetic sequence. GOOD ====Solution==== Let $a_1$ be first term and $d$ be common difference of A.P. As given \begin{align} &a_6+a_4=6 \\ \implies & a_1+
- Question 14 Exercise 4.2
- and 41. GOOD ====Solution==== Let $A_1, A_2, A_3$ be three arithmetic means between 6 and 41. Then $6,... 2. GOOD ====Solution==== Let $A_1, A_2, A_3, A_4$ be four arithmetic means between 17 and 32.\\ then $
- Question 16 Exercise 4.2
- OD ====Solution==== Let $A_1, A_2, A_3, A_4, A_5$ be five arithmetic means between $5$ and $8$. Then $... 15}{2}\\ &=\dfrac{65}{2}---(i)\end{align} Let $A$ be arithmetic mean between 5 and 8. Then $$A=\dfrac{
- Question 13 & 14 Exercise 4.3
- $50$ so that the sum of the resulting series will be $459$ . ====Solution==== We know that:\\ \begin{a... $50$ as terms,\\ so therefore there would need to be $16$ terms between $1$ and $50$. ====Go To==== <
- Question 9 Exercise 4.4
- ==Solution==== Let $G_1, G_2, G_3, G_4$ and $G_5$ be the five geometric means between $\dfrac{32}{9}$ ... ution==== Let $G_1, G_2, G_3, G_4, G_5$ and $G_6$ be the six geometric means between $14$ and $-\dfrac
- Question 1 and 2 Exercise 4.1
- on==== It is infinite sequence, the last term may be $0$ or $1$ , but how much terms in this sequence,
- Question 17 Exercise 4.2
- ====Solution==== Let $A_1, A_2, A_3, \ldots, A_n$ be $n$ arithmetic means between 5 and 32. Then $5, A
- Question 2 Exercise 4.3
- or } n=-\dfrac{50}{8}\end{align} Since $n$ cannot be negative or in fraction, thus $n=9$. Now $$a_9=a_
- Question 9 & 10 Exercise 4.3
- ext { and } a_n=693 .$$\\ Let the number of terms be $n$. Then\\ \begin{align}a_n&=a_1+(n-1) d \text {
- Question 10 Exercise 4.4
- .M by $18$ . ====Solution==== Let the two numbers be $a$ and $b$ \\ Condition-$1$\\ The difference bet
- Question 11 Exercise 4.4
- ====Solution==== Let $G_1, G_2, G_9, \ldots, G_n$ be the $n$ geometric means between $a$ and $b$,\\ th