Search
You can find the results of your search below.
Fulltext results:
- Question 3 & 4 Exercise 4.3
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... tion 3===== Find sum of all the numbers divisible by $5$ from $25$ to $350$. GOOD ====Solution==== The numbers divisible by $5$ from $25$ tò $350$ are\\ $$25,30,35, \ldots, ... pose the three numbers are $a-d, a, a+d$\\. then by first condition their sum is equal to $36$\\ \beg
- Question 3 and 4 Exercise 4.1
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... te down the nth term of the sequence as suggested by the pattern. $\dfrac{1}{2}, \dfrac{2}{3} \dfrac{3... te down the nth term of the sequence as suggested by the pattern. $2,-4,6,-8,10, \ldots$ ====Solution=... te down the nth term of the sequence as suggested by the pattern. $1,-1,1,-1, \ldots$ ====Solution====
- Question 6 Exercise 4.1
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... tion 6(i)===== Find the Pascal sequence for $n=5$ by using its general recursive definition. GOOD ===... ion 6(ii)===== Find the Pascal sequence for $n=6$ by using its general recursive definition. ====Solut... on 6(iii)===== Find the Pascal sequence for $n=8$ by using its general recursive definition. =====Solu
- Question 10 Exercise 4.4
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... ween them is $48$ and their A.M exceeds their G.M by $18$ . ====Solution==== Let the two numbers be $a... {2}$$ Condition-$2$\\ Their A.M exceeds their G.M by 18.\\ Therefore $A \cdot M=G \cdot M+18$ or $A \c... ac{a+b}{2}-\sqrt{a b}=18$$ Multiplying both sides by 2\\ $$(a+b)-2 \sqrt{a b}=36 \text {. }$$ From (i)
- Question 4 Exercise 4.5
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... =0.01<1$.\\ Therefore the sum exists and is given by\\ \begin{align}S_{\infty}&=\dfrac{a_1}{1-r}\\ &=\... Thus the sum of the given series exists and given by $$S_{\infty}=\dfrac{a_1}{1-r}$$,\\ putting $a_1, ... 01<1 \end{align} Thus the sum exists and is given by\\ \begin{align}S_{\infty}&=\dfrac{a_1}{1-r}\end{a
- Question 8 Exercise 4.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... -\dfrac{2(S-b)}{b}\end{align} Dividing both sides by $2$\\ \begin{align}\dfrac{S-b}{b}-\dfrac{S-a}{a}&... frac{(b-c) S}{b c}\end{align} Dividing both sides by $S$\\ \begin{align}\Rightarrow \dfrac{a-b}{a b}&=
- Question 2 & 3 Exercise 4.4
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... 2=27\\ a_5&=a_1 r^4=243.\end{align} Dividing (ii) by (i), we get\\ \begin{align}\dfrac{a_1 r^4}{a_1 r^... a_1 r^2=-\sqrt{2}...(ii)\end{align} Dividing (ii) by (i), we get\\ \begin{align}\dfrac{a_1 r^2}{a_1 r}
- Question 3 and 4 Exercise 4.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... estion 4===== The $n$th term of sequence is given by $a_n=2n+7$. Show that it is an arithmetic progres
- Question 9 Exercise 4.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... 4+7(-3)=3. \end{align} Hence the distance covered by the ball in 8th second is $3m$. GOOD ====Go To==
- Question 10 Exercise 4.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... on 10===== The population of a town is decreasing by $500$ inhabitants each year If its population in
- Question 11 Exercise 4.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... on==== Suppose $a_1$ represent the distance climb by Ahmad and Akram in first hour. Then $$a_1=1000.$$
- Question 16 Exercise 4.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... 2}$$ Multiplying both side of the arithmetic mean by 5 , we get\\ $$5 A=\dfrac{65}{2}---(ii)$$ From (i
- Question 17 Exercise 4.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... 2}=32.$$ We know $a_n=a_1+(n-1) d$, replacing $n$ by $n+2$, we get\\ \begin{align}a_{n+2}&=a_1+(n+2-1)
- Question 7 & 8 Exercise 4.3
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... -5=6$ then sum of the $n$ terms,\\ let say denote by $S_n^{\prime \prime}$, then\\ $$S_n^{\prime \prim
- Question 11 & 12 Exercise 4.3
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... =32$.\\ The total distance in six second is $S_6$ By arithmetic sum formula, we know\\ \begin{align}S_