Samacheer Kalvi Class 11 Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Exercise 5.2

 

Question 1. Write the first 6 terms of the sequence whose nth terms are given below and classify them as Arithmetic progression Geometric progression, Arithmetic - geometric progression, Harmonic progression and none of them.
(i) \( \frac{1}{2^{n+1}} \)
(ii) \( \frac{(n + 1) (n + 2)}{(n + 3) (n + 4)} \)
(iii) \( 4\left(\frac{1}{2}\right)^{n} \)
(iv) \( \frac{(-1)^{n}}{n} \)
(v) \( \frac{2 n+3}{3 n+4} \)
(vi) 2018
(vii) \( \frac{3 n-2}{3^{n-1}} \)
Answer:
(i) Given \( a_n = \frac{1}{2^{n+1}} \).
Let's find the first 6 terms by putting n = 1, 2, 3, 4, 5, 6:
\( a_1 = \frac{1}{2^{1+1}} = \frac{1}{2^2} = \frac{1}{4} \)
\( a_2 = \frac{1}{2^{2+1}} = \frac{1}{2^3} = \frac{1}{8} \)
\( a_3 = \frac{1}{2^{3+1}} = \frac{1}{2^4} = \frac{1}{16} \)
\( a_4 = \frac{1}{2^{4+1}} = \frac{1}{2^5} = \frac{1}{32} \)
\( a_5 = \frac{1}{2^{5+1}} = \frac{1}{2^6} = \frac{1}{64} \)
\( a_6 = \frac{1}{2^{6+1}} = \frac{1}{2^7} = \frac{1}{128} \)
The sequence is \( \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \frac{1}{64}, \frac{1}{128}, \dots \)
To check if it's an A.P., we look for a common difference:
\( a_2 - a_1 = \frac{1}{8} - \frac{1}{4} = \frac{1-2}{8} = -\frac{1}{8} \)
\( a_3 - a_2 = \frac{1}{16} - \frac{1}{8} = \frac{1-2}{16} = -\frac{1}{16} \)
Since the differences are not the same, it is not an A.P.
To check if it's a G.P., we look for a common ratio:
\( \frac{a_2}{a_1} = \frac{1/8}{1/4} = \frac{1}{8} \times 4 = \frac{1}{2} \)
\( \frac{a_3}{a_2} = \frac{1/16}{1/8} = \frac{1}{16} \times 8 = \frac{1}{2} \)
Since the ratio is constant \( (\frac{1}{2}) \), the sequence is a Geometric Progression (G.P.) with first term \( a = \frac{1}{4} \) and common ratio \( r = \frac{1}{2} \).
In simple words: We find the first six numbers by putting n=1 to 6 in the given formula. Then we check if the difference between terms is always the same (A.P.) or if the ratio between terms is always the same (G.P.). In this case, the ratio is always \( \frac{1}{2} \), so it's a Geometric Progression.

(ii) Given \( a_n = \frac{(n + 1) (n + 2)}{(n + 3) (n + 4)} \).
Let's find the first 6 terms:
\( a_1 = \frac{(1+1)(1+2)}{(1+3)(1+4)} = \frac{2 \times 3}{4 \times 5} = \frac{6}{20} = \frac{3}{10} \)
\( a_2 = \frac{(2+1)(2+2)}{(2+3)(2+4)} = \frac{3 \times 4}{5 \times 6} = \frac{12}{30} = \frac{2}{5} \)
\( a_3 = \frac{(3+1)(3+2)}{(3+3)(3+4)} = \frac{4 \times 5}{6 \times 7} = \frac{20}{42} = \frac{10}{21} \)
\( a_4 = \frac{(4+1)(4+2)}{(4+3)(4+4)} = \frac{5 \times 6}{7 \times 8} = \frac{30}{56} = \frac{15}{28} \)
\( a_5 = \frac{(5+1)(5+2)}{(5+3)(5+4)} = \frac{6 \times 7}{8 \times 9} = \frac{42}{72} = \frac{7}{12} \)
\( a_6 = \frac{(6+1)(6+2)}{(6+3)(6+4)} = \frac{7 \times 8}{9 \times 10} = \frac{56}{90} = \frac{28}{45} \)
The sequence is \( \frac{3}{10}, \frac{2}{5}, \frac{10}{21}, \frac{15}{28}, \frac{7}{12}, \frac{28}{45}, \dots \)
To check for A.P.:
\( a_2 - a_1 = \frac{2}{5} - \frac{3}{10} = \frac{4-3}{10} = \frac{1}{10} \)
\( a_3 - a_2 = \frac{10}{21} - \frac{2}{5} = \frac{50-42}{105} = \frac{8}{105} \)
The differences are not constant, so it's not an A.P.
To check for G.P.:
\( \frac{a_2}{a_1} = \frac{2/5}{3/10} = \frac{2}{5} \times \frac{10}{3} = \frac{4}{3} \)
\( \frac{a_3}{a_2} = \frac{10/21}{2/5} = \frac{10}{21} \times \frac{5}{2} = \frac{25}{21} \)
The ratios are not constant, so it's not a G.P.
This sequence is neither an Arithmetic Progression (A.P.), a Geometric Progression (G.P.), nor an Arithmetic-Geometric Progression (AGP).
In simple words: We find the first six terms by putting n=1 to 6 in the formula. Then we check if there's a constant difference or a constant ratio between consecutive terms. If neither is found, it means the sequence does not follow these common patterns.

(iii) Given the nth term \( a_n = 4\left(\frac{1}{2}\right)^{n} \).
Let's find the first 6 terms:
\( a_1 = 4\left(\frac{1}{2}\right)^{1} = 4 \times \frac{1}{2} = 2 \)
\( a_2 = 4\left(\frac{1}{2}\right)^{2} = 4 \times \frac{1}{4} = 1 \)
\( a_3 = 4\left(\frac{1}{2}\right)^{3} = 4 \times \frac{1}{8} = \frac{1}{2} \)
\( a_4 = 4\left(\frac{1}{2}\right)^{4} = 4 \times \frac{1}{16} = \frac{1}{4} \)
\( a_5 = 4\left(\frac{1}{2}\right)^{5} = 4 \times \frac{1}{32} = \frac{1}{8} \)
\( a_6 = 4\left(\frac{1}{2}\right)^{6} = 4 \times \frac{1}{64} = \frac{1}{16} \)
The sequence is \( 2, 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dots \)
To check for G.P., we find the common ratio:
\( \frac{a_2}{a_1} = \frac{1}{2} \)
\( \frac{a_3}{a_2} = \frac{1/2}{1} = \frac{1}{2} \)
Since the ratio is constant \( (\frac{1}{2}) \), the sequence is a Geometric Progression (G.P.) with first term \( a=2 \) and common ratio \( r=\frac{1}{2} \). A geometric progression grows or shrinks by multiplying a fixed number each time.
In simple words: We calculate the first six numbers using the given formula. We observe that each number is half of the one before it. This means it has a constant ratio, so it is a Geometric Progression.

(iv) Given the nth term \( a_n = \frac{(-1)^{n}}{n} \).
Let's find the first 6 terms:
\( a_1 = \frac{(-1)^1}{1} = -1 \)
\( a_2 = \frac{(-1)^2}{2} = \frac{1}{2} \)
\( a_3 = \frac{(-1)^3}{3} = -\frac{1}{3} \)
\( a_4 = \frac{(-1)^4}{4} = \frac{1}{4} \)
\( a_5 = \frac{(-1)^5}{5} = -\frac{1}{5} \)
\( a_6 = \frac{(-1)^6}{6} = \frac{1}{6} \)
The sequence is \( -1, \frac{1}{2}, -\frac{1}{3}, \frac{1}{4}, -\frac{1}{5}, \frac{1}{6}, \dots \)
To check for A.P.:
\( a_2 - a_1 = \frac{1}{2} - (-1) = \frac{3}{2} \)
\( a_3 - a_2 = -\frac{1}{3} - \frac{1}{2} = -\frac{2+3}{6} = -\frac{5}{6} \)
The differences are not constant, so it's not an A.P.
To check for G.P.:
\( \frac{a_2}{a_1} = \frac{1/2}{-1} = -\frac{1}{2} \)
\( \frac{a_3}{a_2} = \frac{-1/3}{1/2} = -\frac{1}{3} \times 2 = -\frac{2}{3} \)
The ratios are not constant, so it's not a G.P.
This sequence is neither an Arithmetic Progression (A.P.), a Harmonic Progression (H.P.), nor an Arithmetic-Geometric Progression (AGP).
In simple words: We calculate the first six numbers by replacing 'n' with 1 through 6 in the formula. We see that the terms keep changing sign and the values are not forming a simple pattern of adding or multiplying a fixed number. So, it's not a common type of progression.

(v) Given the nth term \( a_n = \frac{2 n+3}{3 n+4} \).
Let's find the first 6 terms:
\( a_1 = \frac{2(1)+3}{3(1)+4} = \frac{2+3}{3+4} = \frac{5}{7} \)
\( a_2 = \frac{2(2)+3}{3(2)+4} = \frac{4+3}{6+4} = \frac{7}{10} \)
\( a_3 = \frac{2(3)+3}{3(3)+4} = \frac{6+3}{9+4} = \frac{9}{13} \)
\( a_4 = \frac{2(4)+3}{3(4)+4} = \frac{8+3}{12+4} = \frac{11}{16} \)
\( a_5 = \frac{2(5)+3}{3(5)+4} = \frac{10+3}{15+4} = \frac{13}{19} \)
\( a_6 = \frac{2(6)+3}{3(6)+4} = \frac{12+3}{18+4} = \frac{15}{22} \)
The sequence is \( \frac{5}{7}, \frac{7}{10}, \frac{9}{13}, \frac{11}{16}, \frac{13}{19}, \frac{15}{22}, \dots \)
To check for A.P.:
\( a_2 - a_1 = \frac{7}{10} - \frac{5}{7} = \frac{49-50}{70} = -\frac{1}{70} \)
\( a_3 - a_2 = \frac{9}{13} - \frac{7}{10} = \frac{90-91}{130} = -\frac{1}{130} \)
The differences are not constant, so it's not an A.P.
To check for G.P.:
\( \frac{a_2}{a_1} = \frac{7/10}{5/7} = \frac{7}{10} \times \frac{7}{5} = \frac{49}{50} \)
\( \frac{a_3}{a_2} = \frac{9/13}{7/10} = \frac{9}{13} \times \frac{10}{7} = \frac{90}{91} \)
The ratios are not constant, so it's not a G.P.
This sequence is neither an Arithmetic Progression (A.P.), a Geometric Progression (G.P.), nor an Arithmetic-Geometric Progression (AGP).
In simple words: We calculate the first six terms by plugging in 'n' values from 1 to 6 into the formula. We then check if the numbers in the sequence are increasing by a fixed amount (A.P.) or by a fixed multiplication factor (G.P.). Since neither pattern fits, it is not one of these types.

(vi) Given the nth term \( a_n = 2018 \).
Let's find the first 6 terms:
\( a_1 = 2018 \)
\( a_2 = 2018 \)<
\( a_3 = 2018 \)
\( a_4 = 2018 \)
\( a_5 = 2018 \)
\( a_6 = 2018 \)
The sequence is \( 2018, 2018, 2018, 2018, 2018, 2018, \dots \)
This is a constant sequence. It has the same common ratio (1) and common difference (0).
Therefore, this sequence is an Arithmetic Progression (A.P.), a Geometric Progression (G.P.), and also an Arithmetic-Geometric Progression (AGP). A constant sequence is a special case that fits all these categories.
In simple words: Every term in this sequence is the same number, 2018. Because the difference between terms is always 0, it's an A.P. Because the ratio between terms is always 1, it's a G.P.

(vii) Given the nth term \( a_n = \frac{3 n-2}{3^{n-1}} \).
Let's find the first 6 terms:
\( a_1 = \frac{3(1)-2}{3^{1-1}} = \frac{1}{3^0} = \frac{1}{1} = 1 \)
\( a_2 = \frac{3(2)-2}{3^{2-1}} = \frac{6-2}{3^1} = \frac{4}{3} \)
\( a_3 = \frac{3(3)-2}{3^{3-1}} = \frac{9-2}{3^2} = \frac{7}{9} \)
\( a_4 = \frac{3(4)-2}{3^{4-1}} = \frac{12-2}{3^3} = \frac{10}{27} \)
\( a_5 = \frac{3(5)-2}{3^{5-1}} = \frac{15-2}{3^4} = \frac{13}{81} \)
\( a_6 = \frac{3(6)-2}{3^{6-1}} = \frac{18-2}{3^5} = \frac{16}{243} \)
The sequence is \( 1, \frac{4}{3}, \frac{7}{9}, \frac{10}{27}, \frac{13}{81}, \frac{16}{243}, \dots \)
The differences between consecutive terms are not constant, so it's not an A.P.
The ratios between consecutive terms are not constant, so it's not a G.P.
This sequence is an Arithmetic-Geometric Progression (AGP). This type of sequence is formed by multiplying terms of an A.P. and a G.P. The numerators (1, 4, 7, 10, 13, 16) form an A.P., and the denominators (1, 3, 9, 27, 81, 243) form a G.P.
In simple words: We find the first six numbers by replacing 'n' in the formula with 1 through 6. We find that neither a constant difference nor a constant ratio exists. However, we can see that the top numbers form an A.P. and the bottom numbers form a G.P., making it an Arithmetic-Geometric Progression.

🎯 Exam Tip: Always calculate at least three terms to verify the type of progression. Two terms are often not enough to distinguish between A.P., G.P., or other sequences.

 

Question 2. Write the first 6 terms of the sequences whose nth term \( a_n \) is given below
(i) \( a_n = \begin{cases} n+1 & \text{if } n \text{ is odd} \\ n & \text{if } n \text{ is even} \end{cases} \)
(ii) \( a_n = \begin{cases} 1 & \text{if } n=1 \\ 2 & \text{if } n=2 \\ a_{n-1} + a_{n-2} & \text{if } n>2 \end{cases} \)
(iii) \( a_n = \begin{cases} n & \text{if } n \text{ is } 1,2 \text{ or } 3 \\ a_{n-1} + a_{n-2} + a_{n-3} & \text{if } n>3 \end{cases} \)
Answer:
(i) Given \( a_n = \begin{cases} n+1 & \text{if } n \text{ is odd} \\ n & \text{if } n \text{ is even} \end{cases} \)
We calculate the first 6 terms using the given conditions:
For \( n=1 \) (odd): \( a_1 = 1+1 = 2 \)
For \( n=2 \) (even): \( a_2 = 2 \)
For \( n=3 \) (odd): \( a_3 = 3+1 = 4 \)
For \( n=4 \) (even): \( a_4 = 4 \)
For \( n=5 \) (odd): \( a_5 = 5+1 = 6 \)
For \( n=6 \) (even): \( a_6 = 6 \)
The first six terms of the sequence are \( 2, 2, 4, 4, 6, 6 \). This sequence shows a clear pattern of repeating terms based on whether 'n' is odd or even.
In simple words: We find the terms by checking if 'n' is an odd or even number. If 'n' is odd, we add 1 to it. If 'n' is even, we just use 'n' as the term. We do this for n=1 to 6.

(ii) Given \( a_n = \begin{cases} 1 & \text{if } n=1 \\ 2 & \text{if } n=2 \\ a_{n-1} + a_{n-2} & \text{if } n>2 \end{cases} \)
We calculate the first 6 terms using the given conditions:
For \( n=1 \): \( a_1 = 1 \)
For \( n=2 \): \( a_2 = 2 \)
For \( n=3 \): \( a_3 = a_{3-1} + a_{3-2} = a_2 + a_1 = 2+1 = 3 \)
For \( n=4 \): \( a_4 = a_{4-1} + a_{4-2} = a_3 + a_2 = 3+2 = 5 \)
For \( n=5 \): \( a_5 = a_{5-1} + a_{5-2} = a_4 + a_3 = 5+3 = 8 \)
For \( n=6 \): \( a_6 = a_{6-1} + a_{6-2} = a_5 + a_4 = 8+5 = 13 \)
The first six terms of the sequence are \( 1, 2, 3, 5, 8, 13 \). This is a Fibonacci-like sequence where each term (after the second) is the sum of the two preceding terms.
In simple words: The first term is 1 and the second is 2. After that, each new term is found by adding the two terms that came right before it. We follow this rule to find the first six numbers.

(iii) Given \( a_n = \begin{cases} n & \text{if } n \text{ is } 1,2 \text{ or } 3 \\ a_{n-1} + a_{n-2} + a_{n-3} & \text{if } n>3 \end{cases} \)
We calculate the first 6 terms using the given conditions:
For \( n=1 \): \( a_1 = 1 \)
For \( n=2 \): \( a_2 = 2 \)
For \( n=3 \): \( a_3 = 3 \)
For \( n=4 \): \( a_4 = a_{4-1} + a_{4-2} + a_{4-3} = a_3 + a_2 + a_1 = 3+2+1 = 6 \)
For \( n=5 \): \( a_5 = a_{5-1} + a_{5-2} + a_{5-3} = a_4 + a_3 + a_2 = 6+3+2 = 11 \)
For \( n=6 \): \( a_6 = a_{6-1} + a_{6-2} + a_{6-3} = a_5 + a_4 + a_3 = 11+6+3 = 20 \)
The first six terms of the sequence are \( 1, 2, 3, 6, 11, 20 \). This type of sequence, where a term depends on multiple previous terms, is called a recurrence relation.
In simple words: The first three terms are 1, 2, and 3. After that, each new term is found by adding the three terms that came right before it. We use this rule to find the first six numbers in the sequence.

🎯 Exam Tip: When terms are defined recursively (like \( a_n = a_{n-1} + a_{n-2} \)), it's crucial to list the initial conditions clearly. Carefully substitute previous terms to avoid errors in calculation.

 

Question 3. Write the nth term of the following sequences.
(i) 2, 2, 4, 4, 6, 6, ..........
(ii) \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \dots \)
(iii) \( \frac{1}{2}, \frac{3}{4}, \frac{5}{6}, \frac{7}{8}, \frac{9}{10}, \dots \)
(iv) 6, 10, 4, 12, 2, 14, 0, 16, -2, ..........
Answer:
(i) Given sequence: 2, 2, 4, 4, 6, 6, ...
Let's observe the pattern for odd and even positions:
The terms at odd positions (1st, 3rd, 5th, ...) are \( a_1=2, a_3=4, a_5=6, \dots \). These terms form an Arithmetic Progression (A.P.) with a first term \( a=2 \) and common difference \( d=2 \). The general term for these positions would be \( a + (\frac{n-1}{2})d = 2 + (\frac{n-1}{2})2 = 2 + n-1 = n+1 \) for odd \( n \).
The terms at even positions (2nd, 4th, 6th, ...) are \( a_2=2, a_4=4, a_6=6, \dots \). These terms form an A.P. with a first term \( a=2 \) and common difference \( d=2 \). The general term for these positions would be \( a + (\frac{n-2}{2})d = 2 + (\frac{n-2}{2})2 = 2 + n-2 = n \) for even \( n \).
So, the nth term \( a_n \) can be written as:
\( a_n = \begin{cases} n+1 & \text{if } n \text{ is odd} \\ n & \text{if } n \text{ is even} \end{cases} \)
In simple words: We look at the numbers in the sequence. If 'n' is an odd number (like 1, 3, 5), the term is 'n+1'. If 'n' is an even number (like 2, 4, 6), the term is just 'n'. We write these rules together to define the nth term.

(ii) Given sequence: \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \dots \)
Let's analyze the numerator and denominator separately:
The terms in the numerator are 1, 2, 3, 4, 5, ... This is an Arithmetic Progression (A.P.) with a first term \( a=1 \) and common difference \( d=1 \). So, the nth term of the numerator is \( T_n = a + (n-1)d = 1 + (n-1)1 = 1+n-1 = n \).
The terms in the denominator are 2, 3, 4, 5, 6, ... This is an A.P. with a first term \( a=2 \) and common difference \( d=1 \). So, the nth term of the denominator is \( T_n = a + (n-1)d = 2 + (n-1)1 = 2+n-1 = n+1 \).
Therefore, the nth term of the given sequence is \( a_n = \frac{n}{n+1} \) for all natural numbers \( n \). This fraction increases as 'n' gets larger, approaching 1.
In simple words: We look at the top numbers (numerator) and bottom numbers (denominator) of the fractions separately. The top numbers are 1, 2, 3, and so on, so the nth top number is 'n'. The bottom numbers are 2, 3, 4, and so on, so the nth bottom number is 'n+1'. We combine these to get the nth term as \( \frac{n}{n+1} \).

(iii) Given sequence: \( \frac{1}{2}, \frac{3}{4}, \frac{5}{6}, \frac{7}{8}, \frac{9}{10}, \dots \)
Let's analyze the numerator and denominator separately:
The terms in the numerator are 1, 3, 5, 7, 9, ... This is an Arithmetic Progression (A.P.) with a first term \( a=1 \) and common difference \( d=2 \). So, the nth term of the numerator is \( T_n = a + (n-1)d = 1 + (n-1)2 = 1+2n-2 = 2n-1 \).
The terms in the denominator are 2, 4, 6, 8, 10, ... This is an A.P. with a first term \( a=2 \) and common difference \( d=2 \). So, the nth term of the denominator is \( T_n = a + (n-1)d = 2 + (n-1)2 = 2+2n-2 = 2n \).
Therefore, the nth term of the given sequence is \( a_n = \frac{2n-1}{2n} \) for all natural numbers \( n \). This fraction also approaches 1 as 'n' gets larger.
In simple words: For the top numbers, we see they are odd numbers: 1, 3, 5, etc. So the nth top number is \( 2n-1 \). For the bottom numbers, they are even numbers: 2, 4, 6, etc. So the nth bottom number is \( 2n \). Putting them together, the nth term is \( \frac{2n-1}{2n} \).

(iv) Given sequence: 6, 10, 4, 12, 2, 14, 0, 16, -2, ...
Let's analyze the terms at odd and even positions separately:
The terms at odd positions (1st, 3rd, 5th, 7th, 9th, ...) are \( a_1=6, a_3=4, a_5=2, a_7=0, a_9=-2, \dots \). These terms form an Arithmetic Progression (A.P.) with a first term \( a=6 \) and common difference \( d=-2 \). The formula for the nth term for an odd position 'n' is \( a_n = 6 + (\frac{n-1}{2})(-2) = 6 - (n-1) = 6 - n + 1 = 7-n \).
The terms at even positions (2nd, 4th, 6th, 8th, ...) are \( a_2=10, a_4=12, a_6=14, a_8=16, \dots \). These terms form an A.P. with a first term \( a=10 \) and common difference \( d=2 \). The formula for the nth term for an even position 'n' is \( a_n = 10 + (\frac{n-2}{2})(2) = 10 + (n-2) = 10 + n - 2 = 8+n \).
Combining these, the nth term \( a_n \) is defined as:
\( a_n = \begin{cases} 7-n & \text{if } n \text{ is odd} \\ 8+n & \text{if } n \text{ is even} \end{cases} \)
In simple words: We look at the sequence terms based on their position. For odd positions (1st, 3rd, etc.), the pattern is \( 7-n \). For even positions (2nd, 4th, etc.), the pattern is \( 8+n \). This gives us two rules depending on whether 'n' is odd or even.

🎯 Exam Tip: For sequences with alternating patterns, separate the odd-positioned terms and even-positioned terms. Find the rule for each subsequence, then combine them using 'if n is odd' and 'if n is even' conditions.

 

Question 4. The product of three increasing numbers in a G.P is 5832. If we add 6 to the second number and 9 to the third number, then resulting numbers form an A.P. Find the numbers in G.P.
Answer:
Let the three increasing numbers in a Geometric Progression (G.P.) be \( \frac{a}{r}, a, ar \).
The product of these numbers is given as 5832.
\( \frac{a}{r} \times a \times ar = 5832 \)
\( a^3 = 5832 \)
To find 'a', we take the cube root of 5832. We know that \( 18^3 = 5832 \).
So, \( a = 18 \).
Now, according to the problem, if we add 6 to the second number and 9 to the third number, the resulting numbers form an Arithmetic Progression (A.P.).
The new numbers are \( \frac{a}{r}, a+6, ar+9 \).
Since these numbers are in A.P., the middle term is the average of the first and third terms. So, \( 2(a+6) = \frac{a}{r} + (ar+9) \).
Substitute the value of \( a=18 \) into this equation:
\( 2(18+6) = \frac{18}{r} + (18r+9) \)
\( 2(24) = \frac{18}{r} + 18r+9 \)
\( 48 = \frac{18}{r} + 18r+9 \)
Subtract 9 from both sides:
\( 39 = \frac{18}{r} + 18r \)
Multiply the entire equation by \( r \) to remove the fraction (assuming \( r \ne 0 \)):
\( 39r = 18 + 18r^2 \)
Rearrange the terms to form a quadratic equation:
\( 18r^2 - 39r + 18 = 0 \)
Divide by 3 to simplify the equation:
\( 6r^2 - 13r + 6 = 0 \)
Factorize the quadratic equation. We need two numbers that multiply to \( 6 \times 6 = 36 \) and add up to -13. These numbers are -9 and -4.
\( 6r^2 - 9r - 4r + 6 = 0 \)
\( 3r(2r-3) - 2(2r-3) = 0 \)
\( (2r-3)(3r-2) = 0 \)
This gives two possible values for \( r \):
\( 2r-3 = 0 \implies 2r = 3 \implies r = \frac{3}{2} \)
\( 3r-2 = 0 \implies 3r = 2 \implies r = \frac{2}{3} \)
We are looking for three *increasing* numbers in G.P. So, the common ratio \( r \) must be greater than 1.
Case (i): When \( a=18 \) and \( r = \frac{3}{2} \)
The numbers are: \( \frac{18}{3/2} = 18 \times \frac{2}{3} = 12 \)
\( a = 18 \)
\( ar = 18 \times \frac{3}{2} = 27 \)
The numbers are 12, 18, 27. These are increasing.
Case (ii): When \( a=18 \) and \( r = \frac{2}{3} \)
The numbers are: \( \frac{18}{2/3} = 18 \times \frac{3}{2} = 27 \)
\( a = 18 \)
\( ar = 18 \times \frac{2}{3} = 12 \)
The numbers are 27, 18, 12. These are decreasing.
Since the problem specifies *increasing* numbers, the correct set of numbers is 12, 18, 27.
In simple words: We set up the three G.P. numbers using 'a' and 'r'. From their product, we found 'a'. Then we used the condition that when we change the second and third numbers, they form an A.P. This gave us an equation to find 'r'. We found two possible 'r' values. Since the numbers must be increasing, we picked the 'r' that made the sequence go up.

🎯 Exam Tip: When given conditions for both G.P. and A.P. for the same numbers, represent the G.P. terms as \( \frac{a}{r}, a, ar \) to simplify the product. Always check all conditions, like "increasing numbers," at the end to select the correct solution.

 

Question 5. Write the nth term of the sequence \( \frac{3}{1^{2} \cdot 2^{2}}, \frac{5}{2^{2} \cdot 3^{2}}, \frac{7}{3^{2} \cdot 4^{2}}, ............. \) as a difference of two terms.
Answer:
Let the given sequence be \( a_n \). We need to find its nth term and express it as a difference of two terms.
First, let's look at the numerators of the terms: 3, 5, 7, ...
This is an Arithmetic Progression (A.P.) with a first term \( a = 3 \) and a common difference \( d = 5-3 = 2 \).
The nth term of this A.P. is \( T_n = a + (n-1)d = 3 + (n-1)2 = 3 + 2n - 2 = 2n + 1 \).
Next, let's look at the denominators: \( 1^2 \cdot 2^2, 2^2 \cdot 3^2, 3^2 \cdot 4^2, \dots \)
For the 1st term, it's \( 1^2 \cdot 2^2 \).
For the 2nd term, it's \( 2^2 \cdot 3^2 \).
For the 3rd term, it's \( 3^2 \cdot 4^2 \).
So, for the nth term, the denominator will be \( n^2 \cdot (n+1)^2 \).
Combining the numerator and denominator, the nth term of the sequence is:
\( a_n = \frac{2n+1}{n^2(n+1)^2} \)
Now, we need to express this nth term as a difference of two terms.
We can observe that the numerator \( 2n+1 \) can be related to \( (n+1)^2 - n^2 \).
\( (n+1)^2 - n^2 = (n^2 + 2n + 1) - n^2 = 2n + 1 \).
So, we can replace the numerator \( (2n+1) \) with \( (n+1)^2 - n^2 \).
\( a_n = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} \)
Now, we can split this fraction into two parts:
\( a_n = \frac{(n+1)^2}{n^2(n+1)^2} - \frac{n^2}{n^2(n+1)^2} \)
Simplify each part:
\( a_n = \frac{1}{n^2} - \frac{1}{(n+1)^2} \)
This is the nth term of the sequence expressed as a difference of two terms. This method is often useful for telescoping series.
In simple words: First, we find the formula for the top part (numerator) and the bottom part (denominator) of each fraction. The numerator is \( 2n+1 \) and the denominator is \( n^2 \times (n+1)^2 \). Then, we notice that \( 2n+1 \) is the same as \( (n+1)^2 - n^2 \). We replace the numerator with this, then split the fraction into two simpler ones, which gives us the answer as a subtraction.

🎯 Exam Tip: When asked to express a term as a difference of two terms, especially with squared denominators, look for ways to use the difference of squares formula, \( x^2 - y^2 = (x-y)(x+y) \), or directly expand and simplify terms related to the denominator.

 

Question 6. If \( t_k \) is the \( k^{th} \) term of a G.P then show that \( t_{n-k}, t_n, t_{n+k} \) also form a G.P for any positive integer k.
Answer:
Let the first term of the Geometric Progression (G.P.) be \( A \) and the common ratio be \( R \).
The general formula for the \( m^{th} \) term of a G.P. is \( t_m = AR^{m-1} \).
Using this formula, we can write the terms \( t_{n-k}, t_n, \) and \( t_{n+k} \):
\( t_{n-k} = AR^{(n-k)-1} \)
\( t_n = AR^{n-1} \)
\( t_{n+k} = AR^{(n+k)-1} \)
For three terms to be in G.P., the ratio of consecutive terms must be equal. That is, \( \frac{t_n}{t_{n-k}} = \frac{t_{n+k}}{t_n} \).
Let's find the first ratio, \( \frac{t_n}{t_{n-k}} \):
\( \frac{t_n}{t_{n-k}} = \frac{AR^{n-1}}{AR^{(n-k)-1}} \)
We can cancel out \( A \) and subtract the exponents of \( R \):
\( = R^{(n-1) - ((n-k)-1)} \)
\( = R^{n-1 - n + k + 1} \)
\( = R^k \) (Equation 1)
Now, let's find the second ratio, \( \frac{t_{n+k}}{t_n} \):
\( \frac{t_{n+k}}{t_n} = \frac{AR^{(n+k)-1}}{AR^{n-1}} \)
Again, cancel out \( A \) and subtract the exponents of \( R \):
\( = R^{((n+k)-1) - (n-1)} \)
\( = R^{n+k-1 - n + 1} \)
\( = R^k \) (Equation 2)
From Equation 1 and Equation 2, we see that \( \frac{t_n}{t_{n-k}} = R^k \) and \( \frac{t_{n+k}}{t_n} = R^k \).
Since both ratios are equal to \( R^k \), the terms \( t_{n-k}, t_n, t_{n+k} \) form a Geometric Progression. This shows that terms equidistant from a middle term in a G.P. also maintain the G.P. property.
In simple words: We write down the general formula for any term in a G.P. Then we use this formula for the three terms given: \( t_{n-k} \), \( t_n \), and \( t_{n+k} \). To prove they form a G.P., we check if the ratio between the second and first term is the same as the ratio between the third and second term. Both ratios turn out to be \( R^k \), which confirms they are in a G.P.

🎯 Exam Tip: Remember the general term for a G.P., \( t_m = AR^{m-1} \). When proving properties, clearly write out each term and use exponent rules for simplification. The key property for a G.P. is a constant common ratio.

 

Question 7. If a, b, c are in geometric progression and if \( a^{1/x} = b^{1/y} = c^{1/z} \) are in Arithmetic progression.
Answer:
Given that a, b, c are in Geometric Progression (G.P.).
This means that the square of the middle term equals the product of the other two terms:
\( b^2 = ac \) (Equation 1)
Also given that \( a^{1/x} = b^{1/y} = c^{1/z} \). Let this common value be \( K \).
So, \( a^{1/x} = K \implies a = K^x \)
\( b^{1/y} = K \implies b = K^y \)
\( c^{1/z} = K \implies c = K^z \)
Now, substitute these expressions for a, b, c into Equation 1:
\( (K^y)^2 = K^x \cdot K^z \)
Using the exponent rules \( (X^m)^n = X^{mn} \) and \( X^m \cdot X^n = X^{m+n} \):
\( K^{2y} = K^{x+z} \)
Since the bases are equal, their exponents must also be equal:
\( 2y = x+z \)
This is the condition for x, y, z to be in Arithmetic Progression (A.P.). An A.P. is a sequence where the difference between consecutive terms is constant. This shows a link between geometric properties of a, b, c and arithmetic properties of x, y, z.
In simple words: We start with 'a, b, c' being in G.P., which means \( b^2 = ac \). We are also given that \( a^{1/x}, b^{1/y}, c^{1/z} \) are all equal to some number, let's call it K. This lets us write 'a, b, c' as powers of K. When we put these back into the G.P. rule \( b^2 = ac \) and simplify, we find that \( 2y = x+z \). This equation means that 'x, y, z' are in A.P.

🎯 Exam Tip: When dealing with sequences and powers, often setting a common base or a common value (like K here) simplifies the problem. Remember the core definitions of A.P. (common difference) and G.P. (common ratio).

 

Question 8. The A.M of two numbers exceeds their G.M by 10 and H.M by 16. Find the numbers.
Answer:
Let the two numbers be \( A \) and \( B \).
Let their Arithmetic Mean (A.M.) be \( A_m \), Geometric Mean (G.M.) be \( G_m \), and Harmonic Mean (H.M.) be \( H_m \).
The formulas are:
\( A_m = \frac{A+B}{2} \)
\( G_m = \sqrt{AB} \)
\( H_m = \frac{2AB}{A+B} \)
From the problem statement:
1. The A.M. exceeds the G.M. by 10:
\( A_m = G_m + 10 \implies G_m = A_m - 10 \) (Equation 1)
2. The A.M. exceeds the H.M. by 16:
\( A_m = H_m + 16 \implies H_m = A_m - 16 \) (Equation 2)
We know a very important relationship between A.M., G.M., and H.M. for two positive numbers: \( G_m^2 = A_m \cdot H_m \).
Substitute Equation 1 and Equation 2 into this relationship:
\( (A_m - 10)^2 = A_m (A_m - 16) \)
Expand both sides of the equation:
\( A_m^2 - 2 \cdot A_m \cdot 10 + 10^2 = A_m^2 - 16A_m \)
\( A_m^2 - 20A_m + 100 = A_m^2 - 16A_m \)
Subtract \( A_m^2 \) from both sides:
\( -20A_m + 100 = -16A_m \)
Add \( 20A_m \) to both sides:
\( 100 = 4A_m \)
Divide by 4 to find \( A_m \):
\( A_m = \frac{100}{4} = 25 \)
Now that we have \( A_m \), we can find \( G_m \) and \( H_m \) using Equation 1 and Equation 2:
\( G_m = A_m - 10 = 25 - 10 = 15 \)
\( H_m = A_m - 16 = 25 - 16 = 9 \)
Now we use the definitions of A.M. and G.M. to find the numbers \( A \) and \( B \).
From A.M.: \( \frac{A+B}{2} = 25 \implies A+B = 50 \) (Equation 3)
From G.M.: \( \sqrt{AB} = 15 \implies AB = 15^2 \implies AB = 225 \) (Equation 4)
We have two equations for \( A \) and \( B \). We can form a quadratic equation whose roots are \( A \) and \( B \). The general form is \( x^2 - (A+B)x + AB = 0 \).
Substitute values from Equation 3 and Equation 4:
\( x^2 - 50x + 225 = 0 \)
We need to find two numbers that multiply to 225 and add up to 50. These numbers are 5 and 45.
So, \( (x-5)(x-45) = 0 \)
This gives \( x=5 \) or \( x=45 \).
Therefore, the two numbers are 5 and 45. It's good to check these values: A.M. = \( \frac{5+45}{2} = 25 \); G.M. = \( \sqrt{5 \times 45} = \sqrt{225} = 15 \); H.M. = \( \frac{2 \times 5 \times 45}{5+45} = \frac{450}{50} = 9 \). These fit the conditions.
In simple words: We are given how the Average Mean (A.M.) relates to the Geometric Mean (G.M.) and Harmonic Mean (H.M.). We use a special rule that says G.M. squared is A.M. times H.M. With this rule, we can find the value of A.M. Once we have the A.M., we find G.M. and H.M. Then, we use the formulas for A.M. and G.M. to set up two equations for the two unknown numbers. Solving these equations (which often involves a quadratic equation) gives us the numbers.

🎯 Exam Tip: Always remember the key relationship \( G_m^2 = A_m \cdot H_m \) for two numbers. This property is vital for solving problems where all three means are involved. Once you find \( A_m \) and \( G_m \), use \( A+B = 2A_m \) and \( AB = G_m^2 \) to solve for the numbers.

 

Question 9. If the roots of the equation \( (q – r)x² + (r – p)x + (p – q) = 0 \) are equal then show that p, q, r are in A. P.
Answer:
Given the quadratic equation: \( (q – r)x^2 + (r – p)x + (p – q) = 0 \).
For a quadratic equation \( ax^2 + bx + c = 0 \), the roots are equal if its discriminant is zero, i.e., \( b^2 - 4ac = 0 \).
In our given equation, let's identify a, b, and c:
\( a = (q – r) \)
\( b = (r – p) \)
\( c = (p – q) \)
Now, substitute these into the discriminant condition \( b^2 - 4ac = 0 \):
\( (r – p)^2 - 4(q – r)(p – q) = 0 \)
Expand the terms:
\( (r^2 - 2pr + p^2) - 4(qp - q^2 - rp + rq) = 0 \)
\( r^2 - 2pr + p^2 - 4qp + 4q^2 + 4rp - 4rq = 0 \)
Rearrange and group the terms:
\( p^2 + 4q^2 + r^2 + 2pr - 4pq - 4qr = 0 \)
This expression looks like the expansion of a trinomial square \( (x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx \).
We can rewrite the equation as:
\( p^2 + (-2q)^2 + r^2 + 2(p)(r) + 2(p)(-2q) + 2(-2q)(r) = 0 \)
This matches the form \( (p - 2q + r)^2 = 0 \).
Taking the square root of both sides:
\( p - 2q + r = 0 \)
Rearrange the terms:
\( p + r = 2q \)
This is the condition for three numbers p, q, r to be in an Arithmetic Progression (A.P.). This means the middle term (q) is the arithmetic mean of the other two (p and r).
In simple words: We know that if a quadratic equation has equal answers (roots), then a special part of its formula, called the discriminant, must be zero. We use this rule by putting the parts of our equation into the discriminant formula. After expanding and simplifying, the equation turns into \( (p - 2q + r)^2 = 0 \). This means \( p - 2q + r \) must be zero, which can be rearranged to \( p + r = 2q \). This is the exact rule for numbers to be in an Arithmetic Progression.

🎯 Exam Tip: For problems involving equal roots of a quadratic equation, immediately apply the discriminant condition \( b^2 - 4ac = 0 \). Be careful with algebraic expansions and simplifications, especially when dealing with multiple variables. Recognizing the pattern of a perfect square trinomial can save time.

 

Question 10. If a, b, c are respectively the \( p^{th}, q^{th} \) and \( r^{th} \) terms of a G.P show that \( (q – r) \log a + (r – p) \log b + (p – q) \log c = 0 \)
Answer:
Let the first term of the Geometric Progression (G.P.) be \( A \) and the common ratio be \( R \).
The general formula for the \( n^{th} \) term of a G.P. is \( T_n = AR^{n-1} \).
Given that \( a \) is the \( p^{th} \) term, \( b \) is the \( q^{th} \) term, and \( c \) is the \( r^{th} \) term:
\( a = AR^{p-1} \)
\( b = AR^{q-1} \)
\( c = AR^{r-1} \)
Now, let's take the logarithm of each term. We can use any base for the logarithm (e.g., base 10 or natural logarithm).
\( \log a = \log(AR^{p-1}) = \log A + \log(R^{p-1}) = \log A + (p-1)\log R \) (Equation 1)
\( \log b = \log(AR^{q-1}) = \log A + \log(R^{q-1}) = \log A + (q-1)\log R \) (Equation 2)
\( \log c = \log(AR^{r-1}) = \log A + \log(R^{r-1}) = \log A + (r-1)\log R \) (Equation 3)
We need to show that \( (q – r) \log a + (r – p) \log b + (p – q) \log c = 0 \).
Substitute Equations 1, 2, and 3 into the expression:
\( (q – r)[\log A + (p-1)\log R] + (r – p)[\log A + (q-1)\log R] + (p – q)[\log A + (r-1)\log R] \)
Now, let's group the terms involving \( \log A \) and \( \log R \):
Terms with \( \log A \):
\( (q – r)\log A + (r – p)\log A + (p – q)\log A \)
\( = (q – r + r – p + p – q)\log A \)
\( = (0)\log A = 0 \)
Terms with \( \log R \):
\( (q – r)(p-1)\log R + (r – p)(q-1)\log R + (p – q)(r-1)\log R \)
\( = [(q – r)(p-1) + (r – p)(q-1) + (p – q)(r-1)]\log R \)
Let's expand the terms inside the square brackets:
\( (qp - q - rp + r) + (rq - r - pq + p) + (pr - p - qr + q) \)
\( = qp - q - rp + r + rq - r - pq + p + pr - p - qr + q \)
Now, let's cancel out terms:
\( (qp - pq) + (-q + q) + (-rp + pr) + (r - r) + (rq - qr) + (p - p) = 0 \)
So, the sum of terms with \( \log R \) is also \( 0 \cdot \log R = 0 \).
Therefore, the entire expression simplifies to \( 0 + 0 = 0 \).
Thus, we have shown that \( (q – r) \log a + (r – p) \log b + (p – q) \log c = 0 \). This identity highlights a fundamental property relating the terms and positions in a geometric progression when logarithms are applied.
In simple words: We start by writing the general formulas for 'a, b, c' as terms of a G.P. Then we take the logarithm of each term. We put these log expressions into the equation we need to prove. When we group all the \( \log A \) terms and all the \( \log R \) terms separately, we find that both groups add up to zero. This shows that the whole equation equals zero.

🎯 Exam Tip: This type of problem often involves simplifying polynomial expressions. Always remember the logarithm properties: \( \log(xy) = \log x + \log y \) and \( \log(x^n) = n \log x \). Be systematic in expanding and canceling terms to avoid errors.

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