This is beta site. https://beta.mathcity.org/ 2026-06-07T10:00:48+00:00 https://beta.mathcity.org/ https://beta.mathcity.org/_media/logo.png text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p1?rev=1737476038&do=diff Question 1 Review Exercise 5 Solutions of Question 1 of Review Exercise 5 of Unit 05: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 1 $t_n=6 n+5$$t_{n+1}=$$6 n-1$$6 n+11$$6 n+6$$6 n-5$$1+\dfrac{2}{3}+\dfrac{6}{3^2}+\dfrac{10}{3^3}+\dfrac{14}{3^4}+\ldots$$6$$2$$3$$4$$1+2.2+3.2^2+\cdots+100.2^{\prime \prime}$$99.2^{100}$$100.2^{100}$$99.2^{100}+1$$1000.2^{100}$$n^{t h}$$1.2+2.3+3.4+\… text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p2?rev=1737476038&do=diff Question 2 & 3 Review Exercise Solutions of Question 2 & 3 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$n$$1.2+2.3+3.4+\ldots$$n^{\text {th }}$$$a_n=n(n+1)=n^2+n$$\begin{align} \sum_{r=1}^n a_r&=\sum_{r=1}^n r^2+\sum_{r=1}^n r \\ & =\dfrac{n(n+1)(2 n+1)}{6}+\dfrac{n(n+1)}{2} \\ & =\dfrac{n(n+1)}{2}[\dfrac{2 n+1}{3}+1] \\ & =\dfrac{n(n+1)}{2} \cdot … text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p3?rev=1737476038&do=diff Question 4 Review Exercise Solutions of Question 4 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 4 $\dfrac{1}{1.4 .7}+\dfrac{1}{4.7 .10}+\dfrac{1}{7.10 .13}+\ldots$$1,4,7, \ldots$$$a_n=\dfrac{1}{(3 n-2)(3 n+1)(3 n+4)}$$\begin{align} \dfrac{1}{(3 n-2)(3 n+1)(3 n+4)}&=\dfrac{A}{3 n-2}+\dfrac{B}{3 n+1}+\dfrac{C}{3 n+4}\end{align}$(3 n-2)(3 n+… text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p4?rev=1737476038&do=diff Question 5 & 6 Review Exercise Solutions of Question 5 & 6 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$5+12 x+19 x^2+26 x^3+\ldots$$n$\begin{align}S_n&=5+12 x+19 x^2+26 x^3+\cdots+(7 n-2) x^{n-1}...(i)\\ x S_n&=5 x+12 x^2+19 x^3+\cdots+(7 n-9) x^{n-1}+(7 n-1) x^n....(ii)\end{align}\begin{align}(1-x) S_n&=5+(12-5) x+(19-12) x^2+\cdots\\ &+[7 n-2-(… text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p5?rev=1737476038&do=diff Question 7 Review Exercise Solutions of Question 7 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 7(i) $1.2^2+3.3^2+5.4^2+\ldots$$n$$1,3,5, \ldots,(2 n-1)$$2^2, 3^2, 4^2, \ldots,(n+1)^2$\begin{align} & a_n=(2 n-1)(n+1)^2 \\ & a_n=(2 n-1)(n^2+2 n+1) \\ & a_n=2 n^3+3 n^2-1\end{align}\begin{align} \sum_{r=1}^n a_r&=2 \sum_{r=1}^n r^3+\sum_{r=1… text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p6?rev=1737476038&do=diff Question 8 Review Exercise Solutions of Question 8 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 8(i) $n$$n^{t h}$$n^3+3^n.$$n^h$$$a_n=n^3+3^n$$\begin{align}\sum_{r=1}^n a_r&=\sum_{r=1}^n r^3+\sum_{r=1}^n 3^r \\ & =[\dfrac{n(n+1)}{2}]^2+\dfrac{3(3^n-1)}{3-1} \\ & =\dfrac{n^2(n+1)^2}{4}+\dfrac{3}{2}(3^n-1) \end{align}$n$$$S_n=\dfrac{n^2(n+1… text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p7?rev=1737476038&do=diff Question 9 Review Exercise Solutions of Question 9 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 9(i) $n$$3+7+13+21+31+\ldots$\begin{align} & a_2-a_1=7-3=4 \\ & a_3-a_2=13-7=6 \\ & a_4-a_3=21-13=8 \\ & \ldots \quad \ldots \quad \ldots \\ & \ldots \quad \cdots \quad \ldots \\ & a_n-a_{n-1}=(n-1) \text { term of the series } \\ & 4,6,8, \ldo… text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p8?rev=1737476038&do=diff Question 10 Review Exercise Solutions of Question 10 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 10 $n^{\text {th }}$$n$$1+(1+\dfrac{1}{2})+(1+\dfrac{1}{2}+\dfrac{1}{4})+(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8})+\ldots$\begin{align} a_n&=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots+\dfrac{1}{2^{n-1}} \\ a_n&=\dfrac{1[1-(\dfrac{1}{2})…