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        <title>MathCity.org</title>
        <description>Merging man &amp; maths</description>
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        <item>
            <title>Question 23 and 24, Exercise 4.3</title>
            <link>https://www.mathcity.org/math-11-nbf/sol/unit04/ex4-3-p11</link>
            <description>Question 23 and 24, Exercise 4.3

Solutions of Question 23 and 24 of Exercise 4.3 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $$ 14+16+18+...+a_{25}.$$$a_1=14$$d=16-14=2$$n=25$$a_25$$S_25$\begin{align}
a_n&amp;=a_1+(n-1)d\\
\implies a_{25}&amp;= 14+(25-1)(2)\\
&amp;=62.
\end{align}\begin{align}
S_n&amp;=\frac{n}{2}[a_1+a_n]\\
\implies S_{25}&amp; =\frac{25}{2}[14+62]\\
&amp; =25 \t…</description>
            <author>anonymous@undisclosed.example.com (Anonymous)</author>
            <pubDate>Sat, 14 Sep 2024 16:24:52 +0000</pubDate>
        </item>
        <item>
            <title>Question 25 and 26, Exercise 4.3</title>
            <link>https://www.mathcity.org/math-11-nbf/sol/unit04/ex4-3-p12</link>
            <description>Question 25 and 26, Exercise 4.3

Solutions of Question 25 and 26 of Exercise 4.3 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $$ 6000+70,000+...+a_{20}.$$$a_1=6,000$$d=70,000-6,000=64,000$$n=20$$S_n$\begin{align}
S_n&amp;=\frac{n}{2}[2a_1+(n-1)d]\\
\implies S_{20}&amp; =\frac{20}{2}[2(6,000)+(20-1)(64,000)]\\
&amp; =10 \times [12,000+1,216,000]\\
&amp; =12,280,000.
\end{ali…</description>
            <author>anonymous@undisclosed.example.com (Anonymous)</author>
            <pubDate>Sat, 14 Sep 2024 16:25:19 +0000</pubDate>
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        <item>
            <title>Question 9 and 10, Exercise 4.1</title>
            <link>https://www.mathcity.org/math-11-nbf/sol/unit04/ex4-1-p5</link>
            <description>Question 9 and 10, Exercise 4.1

Solutions of Question 9 and 10 of Exercise 4.1 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $n$$a_{10}$$a_{15}$$a_{n}=(-1)^{n}(n+3)$$n$$a_{10}$$a_{15}$$$a_{n}=(-1)^{n+1}(3 n-5).$$$$a_n = (-1)^{n+1}(3n - 5).$$\begin{align*}
a_1 &amp;= (-1)^{1+1}(3(1) - 5) = (1)(3 - 5) = -2 \\
a_2 &amp;= (-1)^{2+1}(3(2) - 5) = (-1)(6 - 5) = -1 \\
a_3 &amp;=…</description>
            <author>anonymous@undisclosed.example.com (Anonymous)</author>
            <pubDate>Sat, 14 Sep 2024 16:42:36 +0000</pubDate>
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