NCERT Solutions Class 11 Maths Sequences and Series Ex 9.2

Understanding Arithmetic Progression & Arithmetic Means

An Arithmetic Progression (AP) is a sequence of numbers in which the difference of any two consecutive terms is constant. This constant difference is called the common difference, denoted by $d$. The first term is denoted by $a$. The general term (or $n$-th term) is given by $a_n = a + (n-1)d$. The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$ or $S_n = \frac{n}{2}[a + l]$, where $l$ is the last term.

In Class 11, we introduce the concept of Arithmetic Mean (AM). Given two numbers $a$ and $b$, we can insert $m$ numbers $A_1, A_2, \dots, A_m$ between them such that $a, A_1, A_2, \dots, A_m, b$ forms an AP. The common difference of this new AP is calculated using the formula:

$d = \frac{b-a}{m+1}$

Each inserted arithmetic mean $A_k$ can then be calculated as $A_k = a + k \cdot d$.

Step-by-Step Guide to Inserting 'm' Arithmetic Means

1. Identify given parameters — Determine the two boundary numbers $a$ (first term) and $b$ (last term), and the number of means $m$ to insert.
2. Calculate the common difference (d) — Use the formula $d = (b - a) / (m + 1)$. Note that the total number of terms in the resulting AP is $m + 2$.
3. Compute the individual means — Calculate each mean $A_k$ using the formula $A_k = a + k \cdot d$ for $k = 1, 2, \dots, m$.

Exam Traps & Smart Tricks

1. Choosing symmetric terms: When you are asked to find three numbers in an AP whose sum is given, always assume the terms are $a-d$, $a$, and $a+d$. This simplifies calculations because adding them eliminates $d$, instantly giving you $3a = \text{sum}$. For four terms, use $a-3d$, $a-d$, $a+d$, $a+3d$ with common difference $2d$.

2. General Term from Sum: If you are given the sum of $n$ terms $S_n$, you can easily find the $n$-th term $a_n$ using the relation $a_n = S_n - S_{n-1}$. This is a high-yield formula for CBSE Class 11 exams.

Practice Questions with Solutions

• Q: Find the sum of odd integers from 1 to 2001. A: Step 1: Identify the sequence of odd integers: 1, 3, 5, ..., 2001. This is an AP with first term $a = 1$, common difference $d = 2$, and last term $l = 2001$. Step 2: Find the number of terms ($n$). $a_n = a + (n-1)d \implies 2001 = 1 + (n-1)2$ $2000 = 2(n-1) \implies n-1 = 1000 \implies n = 1001$. Step 3: Calculate the sum ($S_n$). $S_n = \frac{n}{2}[a + l] = \frac{1001}{2}[1 + 2001] = \frac{1001}{2}[2002] = 1001 \times 1001 = 1,002,001$. Final answer: The sum of odd integers from 1 to 2001 is 1,002,001.
• Q: If the sum of $n$ terms of an AP is $np + \frac{1}{2}n(n-1)q$, where $p$ and $q$ are constants, find the common difference. A: Step 1: Let the sum of $n$ terms be $S_n = np + \frac{1}{2}n(n-1)q$. Step 2: Find the first term ($a_1$) by putting $n = 1$. $S_1 = 1(p) + \frac{1}{2}(1)(0)q = p$. Thus, $a_1 = p$. Step 3: Find the sum of first two terms ($S_2$) by putting $n = 2$. $S_2 = 2p + \frac{1}{2}(2)(1)q = 2p + q$. Step 4: Find the second term ($a_2$). $a_2 = S_2 - S_1 = (2p + q) - p = p + q$. Step 5: Find the common difference ($d$). $d = a_2 - a_1 = (p + q) - p = q$. Final answer: The common difference is $q$.
• Q: The ratio of the sums of $m$ and $n$ terms of an AP is $m^2 : n^2$. Show that the ratio of $m$-th and $n$-th term is $(2m-1) : (2n-1)$. A: Step 1: Write down the ratio of the sums. $\frac{S_m}{S_n} = \frac{\frac{m}{2}[2a + (m-1)d]}{\frac{n}{2}[2a + (n-1)d]} = \frac{m^2}{n^2}$. Step 2: Simplify the equation. $\frac{2a + (m-1)d}{2a + (n-1)d} = \frac{m}{n}$. Step 3: We need to find the ratio of the $m$-th and $n$-th terms, which is $\frac{a_m}{a_n} = \frac{a + (m-1)d}{a + (n-1)d} = \frac{2a + (2m-2)d}{2a + (2n-2)d}$. Step 4: Substitute $m \to 2m-1$ and $n \to 2n-1$ in the equation from Step 2. $\frac{2a + ((2m-1)-1)d}{2a + ((2n-1)-1)d} = \frac{2m-1}{2n-1} \implies \frac{2a + (2m-2)d}{2a + (2n-2)d} = \frac{2m-1}{2n-1} \implies \frac{a_m}{a_n} = \frac{2m-1}{2n-1}$. Final answer: Hence proved, the ratio of the $m$-th and $n$-th term is $(2m-1) : (2n-1)$.
• Q: Insert 6 numbers between 3 and 24 such that the resulting sequence is an AP. A: Step 1: Let the 6 numbers be $A_1, A_2, A_3, A_4, A_5, A_6$. The sequence is $3, A_1, A_2, A_3, A_4, A_5, A_6, 24$. Step 2: Here, $a = 3$, $b = 24$, and $m = 6$ (number of means). Step 3: Calculate the common difference ($d$). $d = \frac{b-a}{m+1} = \frac{24 - 3}{6+1} = \frac{21}{7} = 3$. Step 4: Write down the means. $A_1 = a + d = 3 + 3 = 6$ $A_2 = a + 2d = 3 + 6 = 9$ $A_3 = a + 3d = 3 + 9 = 12$ $A_4 = a + 4d = 3 + 12 = 15$ $A_5 = a + 5d = 3 + 15 = 18$ $A_6 = a + 6d = 3 + 18 = 21$ Final answer: The 6 inserted numbers are 6, 9, 12, 15, 18, and 21.

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