RBSE Solutions Class 11 Maths Chapter 8 Sequence, Progression, and Series Exercise 8.1

Question 1. From the following sequence which are in A.P. ?
(i) 2, 6, 11, 17,...
(ii) 1, 1.4, 1.8, 2.2,...
(iii) -7, -5, -3, -1,...
(iv) 1, 8, 27, 64,...
Answer: To find if a sequence is an Arithmetic Progression (A.P.), we check if the difference between any two consecutive terms is constant. This constant difference is called the common difference.
(i) For the sequence 2, 6, 11, 17,...
The first difference is \( 6 - 2 = 4 \).
The second difference is \( 11 - 6 = 5 \).
Since the differences (4 and 5) are not the same, this sequence is not an A.P.
(ii) For the sequence 1, 1.4, 1.8, 2.2,...
The first difference is \( 1.4 - 1 = 0.4 \).
The second difference is \( 1.8 - 1.4 = 0.4 \).
The third difference is \( 2.2 - 1.8 = 0.4 \).
Since the common difference is constant (0.4), this sequence is an A.P.
(iii) For the sequence -7, -5, -3, -1,...
The first difference is \( -5 - (-7) = -5 + 7 = 2 \).
The second difference is \( -3 - (-5) = -3 + 5 = 2 \).
The third difference is \( -1 - (-3) = -1 + 3 = 2 \).
Since the common difference is constant (2), this sequence is an A.P.
(iv) For the sequence 1, 8, 27, 64,...
The first difference is \( 8 - 1 = 7 \).
The second difference is \( 27 - 8 = 19 \).
Since the differences (7 and 19) are not the same, this sequence is not an A.P.
In simple words: We check if the gap between numbers is always the same. If it is, then it's an A.P. For (i) and (iv), the gaps change, so they are not A.P. For (ii) and (iii), the gaps stay the same, so they are A.P.

🎯 Exam Tip: To quickly check if a sequence is an A.P., calculate the difference between the first two terms and the next two terms. If they are not equal, it's definitely not an A.P.

Question 2. Find first term, common difference and 5th term of those sequence which have following nth term :
(i) \( 3n+ 7 \)
(ii) \( a + (n - 1)d \)
(iii) \( 5 - 3n \)
Answer: We can find the terms of a sequence by putting the value of \( n \) (like 1 for first term, 2 for second term, and so on) into the given \( n^{th} \) term formula. Once we have the first two terms, we can find the common difference.
(i) For the sequence with \( n^{th} \) term \( T_n = 3n + 7 \):
First term \( T_1 = 3(1) + 7 = 3 + 7 = 10 \).
Second term \( T_2 = 3(2) + 7 = 6 + 7 = 13 \).
Common difference \( d = T_2 - T_1 = 13 - 10 = 3 \).
Fifth term \( T_5 = 3(5) + 7 = 15 + 7 = 22 \).
So, for this sequence, the first term is 10, the common difference is 3, and the 5th term is 22.
(ii) For the sequence with \( n^{th} \) term \( T_n = a + (n - 1)d \):
This is the general formula for an A.P. Its first term is given by setting \( n=1 \).
First term \( T_1 = a + (1 - 1)d = a + 0 \times d = a \).
Second term \( T_2 = a + (2 - 1)d = a + d \).
Common difference \( d = T_2 - T_1 = (a + d) - a = d \).
Fifth term \( T_5 = a + (5 - 1)d = a + 4d \).
So, for this general A.P. formula, the first term is \( a \), the common difference is \( d \), and the 5th term is \( a + 4d \).
(iii) For the sequence with \( n^{th} \) term \( T_n = 5 - 3n \):
First term \( T_1 = 5 - 3(1) = 5 - 3 = 2 \).
Second term \( T_2 = 5 - 3(2) = 5 - 6 = -1 \).
Common difference \( d = T_2 - T_1 = -1 - 2 = -3 \).
Fifth term \( T_5 = 5 - 3(5) = 5 - 15 = -10 \).
So, for this sequence, the first term is 2, the common difference is -3, and the 5th term is -10.
In simple words: To find the first term, plug \( n=1 \) into the formula. For the second term, plug \( n=2 \). The common difference is the second term minus the first. To find the fifth term, plug \( n=5 \) into the formula.

🎯 Exam Tip: Always remember that the common difference \( d \) for an arithmetic progression (A.P.) defined by \( T_n = An + B \) is simply the coefficient of \( n \), which is \( A \).

Question 3. Show that sequence of following nth terms, are not in A.P. :
(i) \( \frac { n }{ n+1 } \)
(ii) \( n² + 1 \)
Answer: For a sequence to be an Arithmetic Progression (A.P.), the difference between consecutive terms, \( T_{n+1} - T_n \), must be a constant value that does not depend on \( n \). If the difference changes with \( n \), then it is not an A.P.
(i) For the sequence with \( n^{th} \) term \( T_n = \frac { n }{ n+1 } \):
First, we find the \( (n+1)^{th} \) term: \( T_{n+1} = \frac { n+1 }{ (n+1)+1 } = \frac { n+1 }{ n+2 } \).
Now, we calculate the difference \( T_{n+1} - T_n \):
\( T_{n+1} - T_n = \frac { n+1 }{ n+2 } - \frac { n }{ n+1 } \)
\( = \frac { (n+1)(n+1) - n(n+2) }{ (n+2)(n+1) } \)
\( = \frac { (n^2 + 2n + 1) - (n^2 + 2n) }{ (n+1)(n+2) } \)
\( = \frac { n^2 + 2n + 1 - n^2 - 2n }{ (n+1)(n+2) } \)
\( = \frac { 1 }{ (n+1)(n+2) } \)
Since the difference \( \frac { 1 }{ (n+1)(n+2) } \) depends on \( n \) (it changes as \( n \) changes), the sequence is not an A.P.
(ii) For the sequence with \( n^{th} \) term \( T_n = n^2 + 1 \):
First, we find the \( (n+1)^{th} \) term: \( T_{n+1} = (n+1)^2 + 1 = n^2 + 2n + 1 + 1 = n^2 + 2n + 2 \).
Now, we calculate the difference \( T_{n+1} - T_n \):
\( T_{n+1} - T_n = (n^2 + 2n + 2) - (n^2 + 1) \)
\( = n^2 + 2n + 2 - n^2 - 1 \)
\( = 2n + 1 \)
Since the difference \( 2n + 1 \) depends on \( n \) (it changes as \( n \) changes), the sequence is not an A.P.
In simple words: For a sequence to be an A.P., the step from one number to the next must always be the same. We check this by subtracting the \( n^{th} \) term from the \( (n+1)^{th} \) term. If this answer changes when \( n \) changes, then it is not an A.P. Both given sequences show differences that depend on \( n \), so they are not A.P.s.

🎯 Exam Tip: When proving a sequence is not an A.P., showing that \( T_{n+1} - T_n \) contains \( n \) or any variable (and is not a constant) is the key step.

Question 4. In an A.P. 2 +5+8+11+... which term is 65?
Answer: We are given an Arithmetic Progression (A.P.) as 2, 5, 8, 11, ... and we want to find out which term in this sequence is 65. The formula for the \( n^{th} \) term of an A.P. is \( T_n = a + (n-1)d \), where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the term number.
From the given A.P.:
The first term \( a = 2 \).
The common difference \( d = 5 - 2 = 3 \).
We are looking for the term \( n \) such that \( T_n = 65 \).
Using the formula:
\( T_n = a + (n-1)d \)
\( 65 = 2 + (n-1)3 \)
Now, we solve for \( n \):
\( 65 - 2 = (n-1)3 \)
\( 63 = (n-1)3 \)
Divide both sides by 3:
\( \frac { 63 }{ 3 } = n-1 \)
\( 21 = n-1 \)
Add 1 to both sides:
\( n = 21 + 1 \)
\( n = 22 \)
Therefore, the 22nd term of the A.P. is 65.
In simple words: We have a list of numbers where each number increases by 3. We start at 2. We need to find which number in this list is 65. By using the formula for an A.P., we find that 65 is the 22nd number in the sequence.

🎯 Exam Tip: Always identify the first term \( a \) and common difference \( d \) correctly at the start. Then, substitute these values into the \( n^{th} \) term formula and carefully solve for \( n \).

Question 5. In an A.P. 4 + 9 + 14 +19 +... + 124, find 13th term from last.
Answer: We are given an Arithmetic Progression (A.P.) and need to find a specific term starting from the end of the sequence. The formula to find the \( k^{th} \) term from the last of an A.P. is \( T_{last-k+1} = l - (k-1)d \), where \( l \) is the last term, \( k \) is the position from the last, and \( d \) is the common difference.
From the given A.P.: 4, 9, 14, 19, ..., 124
The first term \( a = 4 \).
The common difference \( d = 9 - 4 = 5 \).
The last term \( l = 124 \).
We need to find the 13th term from the last, so \( k = 13 \).
Using the formula for the \( k^{th} \) term from the last:
\( 13^{th} \) term from last \( = l - (k-1)d \)
\( = 124 - (13-1) \times 5 \)
\( = 124 - (12) \times 5 \)
\( = 124 - 60 \)
\( = 64 \)
Therefore, the 13th term from the last in this A.P. is 64.
In simple words: We have a list of numbers that go up by 5 each time, ending at 124. We want to find the 13th number if we count backwards from 124. Starting from 124 and going back 12 steps of 5, we land on 64.

🎯 Exam Tip: Remember the formula for the \( k^{th} \) term from the end: \( l - (k-1)d \). Pay careful attention to the subtraction sign and the order of operations.

Question 6. In an A.P. 2 + 5 + 8 + 11 + ... if last term is 95, then find the number of terms of the series.
Answer: We need to find how many terms are in the given Arithmetic Progression (A.P.) when the first term, common difference, and the last term are known. The formula for the \( n^{th} \) term of an A.P. is \( T_n = a + (n-1)d \), where \( T_n \) is also the last term \( l \).
From the given A.P.: 2, 5, 8, 11, ...
The first term \( a = 2 \).
The common difference \( d = 5 - 2 = 3 \).
The last term \( l = 95 \).
We set the \( n^{th} \) term \( T_n \) equal to the last term \( l \) and solve for \( n \):
\( l = a + (n-1)d \)
\( 95 = 2 + (n-1)3 \)
Now, we solve for \( n \):
\( 95 - 2 = (n-1)3 \)
\( 93 = (n-1)3 \)
Divide both sides by 3:
\( \frac { 93 }{ 3 } = n-1 \)
\( 31 = n-1 \)
Add 1 to both sides:
\( n = 31 + 1 \)
\( n = 32 \)
Thus, there are 32 terms in this A.P.
In simple words: We have a sequence of numbers that starts at 2 and adds 3 each time, ending at 95. We want to count how many numbers are in this list. By using the A.P. formula, we find there are 32 terms.

🎯 Exam Tip: Clearly identify \( a, d, \) and \( l \) first. Then, be careful with the algebraic steps to isolate \( n \), especially when distributing or dividing.

Question 7. If 9th term of A.P. is zero, then prove that 29th term is twice the 19th term.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the Arithmetic Progression (A.P.). The \( n^{th} \) term of an A.P. is given by the formula \( T_n = a + (n-1)d \).
We are given that the 9th term of the A.P. is zero.
So, \( T_9 = a + (9-1)d = 0 \)
\( a + 8d = 0 \)
\( \implies a = -8d \) (Equation 1)
Now, we need to express the 29th term and the 19th term in terms of \( a \) and \( d \), and then use Equation 1.
The 29th term is \( T_{29} = a + (29-1)d = a + 28d \).
Substitute \( a = -8d \) from Equation 1 into the expression for \( T_{29} \):
\( T_{29} = (-8d) + 28d \)
\( T_{29} = 20d \) (Equation 2)
The 19th term is \( T_{19} = a + (19-1)d = a + 18d \).
Substitute \( a = -8d \) from Equation 1 into the expression for \( T_{19} \):
\( T_{19} = (-8d) + 18d \)
\( T_{19} = 10d \) (Equation 3)
Now, we compare Equation 2 and Equation 3:
We see that \( T_{29} = 20d \) and \( T_{19} = 10d \).
This means \( T_{29} = 2 \times (10d) \).
\( \implies T_{29} = 2 \times T_{19} \)
Thus, it is proven that the 29th term is twice the 19th term.
In simple words: We are told that the 9th number in a sequence is zero. We use this to find a connection between the first number and the common step size. Then, we use this connection to write the 29th and 19th numbers in a simpler way. We find that the 29th number is exactly double the 19th number.

🎯 Exam Tip: The key to solving such proofs is to use the given condition (e.g., \( T_9 = 0 \)) to establish a relationship between \( a \) and \( d \), and then substitute this relationship into the terms you need to compare.

Question 8. How many two digit natural number which are divisible by 3 ?
Answer: We need to find the count of all two-digit natural numbers that can be divided by 3 without any remainder. These numbers form an Arithmetic Progression (A.P.) because they have a constant difference of 3 between them.
The smallest two-digit natural number is 10.
The largest two-digit natural number is 99.
Let's find the first two-digit number divisible by 3. Starting from 10, the first number divisible by 3 is 12 (since \( 10 \div 3 \) leaves a remainder, but \( 12 \div 3 = 4 \)).
So, the first term \( a = 12 \).
The numbers are divisible by 3, so the common difference \( d = 3 \).
Now, let's find the last two-digit number divisible by 3. The largest two-digit number is 99, and 99 is exactly divisible by 3 (since \( 99 \div 3 = 33 \)).
So, the last term \( l = 99 \).
We use the \( n^{th} \) term formula for an A.P.: \( l = a + (n-1)d \).
Substitute the values:
\( 99 = 12 + (n-1)3 \)
Now, we solve for \( n \):
\( 99 - 12 = (n-1)3 \)
\( 87 = (n-1)3 \)
Divide both sides by 3:
\( \frac { 87 }{ 3 } = n-1 \)
\( 29 = n-1 \)
Add 1 to both sides:
\( n = 29 + 1 \)
\( n = 30 \)
Thus, there are 30 two-digit natural numbers that are divisible by 3.
In simple words: We want to count all the numbers between 10 and 99 that can be perfectly divided by 3. The first such number is 12, and the last is 99. These numbers form a list where each step is 3. By using the counting formula for such a list, we find that there are 30 such numbers.

🎯 Exam Tip: When dealing with "divisible by" problems, the common difference \( d \) is always the divisor. Always accurately determine the first and last terms within the specified range to avoid errors.

Question 9. If in A.P., pth term is q and qth term is p, then find (p + q)th term.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the Arithmetic Progression (A.P.). The \( n^{th} \) term of an A.P. is given by the formula \( T_n = a + (n-1)d \).
We are given the following information:
The \( p^{th} \) term is \( q \): \( T_p = a + (p-1)d = q \) (Equation 1)
The \( q^{th} \) term is \( p \): \( T_q = a + (q-1)d = p \) (Equation 2)
Our goal is to find the \( (p+q)^{th} \) term, \( T_{p+q} \).
First, subtract Equation 2 from Equation 1 to find \( d \):
\( [a + (p-1)d] - [a + (q-1)d] = q - p \)
\( a + pd - d - a - qd + d = q - p \)
\( pd - qd = q - p \)
\( d(p - q) = -(p - q) \)
\( \implies d = -1 \) (assuming \( p \neq q \))
Now, substitute the value of \( d = -1 \) into Equation 1 to find \( a \):
\( a + (p-1)(-1) = q \)
\( a - (p-1) = q \)
\( a - p + 1 = q \)
\( \implies a = p + q - 1 \)
Now that we have \( a \) and \( d \), we can find the \( (p+q)^{th} \) term:
\( T_{p+q} = a + ((p+q)-1)d \)
Substitute the values of \( a \) and \( d \):
\( T_{p+q} = (p + q - 1) + ((p+q)-1)(-1) \)
\( T_{p+q} = (p + q - 1) - (p+q-1) \)
\( T_{p+q} = 0 \)
Therefore, the \( (p+q)^{th} \) term is 0.
In simple words: In a number sequence, if the \( p^{th} \) number is \( q \) and the \( q^{th} \) number is \( p \), then we can figure out the step size (common difference) and the starting number. Using these, we find that the number at position \( (p+q) \) in the sequence is always 0.

🎯 Exam Tip: This is a standard problem. The trick is to set up two equations based on the given terms, solve them simultaneously to find \( a \) and \( d \), and then use those values to find the required term.

Question 10. In an A.P. pth term is 1/q and qth term is 1/p, then prove that pqth term is unity.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the Arithmetic Progression (A.P.). The \( n^{th} \) term of an A.P. is given by the formula \( T_n = a + (n-1)d \).
We are given the following information:
The \( p^{th} \) term is \( \frac{1}{q} \): \( T_p = a + (p-1)d = \frac{1}{q} \) (Equation 1)
The \( q^{th} \) term is \( \frac{1}{p} \): \( T_q = a + (q-1)d = \frac{1}{p} \) (Equation 2)
Our goal is to prove that the \( pq^{th} \) term, \( T_{pq} \), is unity (i.e., 1).
First, subtract Equation 2 from Equation 1 to find \( d \):
\( [a + (p-1)d] - [a + (q-1)d] = \frac{1}{q} - \frac{1}{p} \)
\( a + pd - d - a - qd + d = \frac{p-q}{pq} \)
\( pd - qd = \frac{p-q}{pq} \)
\( d(p - q) = \frac{p-q}{pq} \)
\( \implies d = \frac{1}{pq} \) (assuming \( p \neq q \))
Now, substitute the value of \( d = \frac{1}{pq} \) into Equation 1 to find \( a \):
\( a + (p-1)\left(\frac{1}{pq}\right) = \frac{1}{q} \)
\( a + \frac{p}{pq} - \frac{1}{pq} = \frac{1}{q} \)
\( a + \frac{1}{q} - \frac{1}{pq} = \frac{1}{q} \)
Subtract \( \frac{1}{q} \) from both sides:
\( a - \frac{1}{pq} = 0 \)
\( \implies a = \frac{1}{pq} \)
Now that we have \( a = \frac{1}{pq} \) and \( d = \frac{1}{pq} \), we can find the \( pq^{th} \) term:
\( T_{pq} = a + ((pq)-1)d \)
Substitute the values of \( a \) and \( d \):
\( T_{pq} = \frac{1}{pq} + (pq-1)\left(\frac{1}{pq}\right) \)
\( T_{pq} = \frac{1}{pq} + \frac{pq}{pq} - \frac{1}{pq} \)
\( T_{pq} = \frac{1}{pq} + 1 - \frac{1}{pq} \)
\( T_{pq} = 1 \)
Thus, it is proven that the \( pq^{th} \) term is unity.
In simple words: We are given an A.P. where the \( p^{th} \) term is \( 1/q \) and the \( q^{th} \) term is \( 1/p \). We use these facts to calculate the first term and the common difference. When we put these values into the formula for the \( pq^{th} \) term, we find that the answer is always 1.

🎯 Exam Tip: This problem is a common variation of the previous one. Be careful with fractions when calculating \( d \) and \( a \), and ensure all algebraic simplifications are correct to arrive at the unity result.