OP Malhotra Class 11 Maths Solutions Chapter 14 Sequence and Series Exercise 14 (B)

Question 1. Write the first six terms of an A.P. in which
(i) a = 5, d = 4;
(ii) a = 98, d = -3;
(iii) a = \( 7\frac { 1 }{ 2 } \), d = \( 1\frac { 1 }{ 2 } \);
Answer:
(i) Given that the first term \( a = 5 \) and the common difference \( d = 4 \).
We know the formula for the nth term of an A.P. is \( T_n = a + (n - 1)d \).
So, we can find the general term \( T_n \):
\( T_n = 5 + (n - 1)4 \)
\( T_n = 5 + 4n - 4 \)
\( T_n = 4n + 1 \)
Now, we find the first six terms:
\( T_1 = 4(1) + 1 = 5 \)
\( T_2 = 4(2) + 1 = 9 \)
\( T_3 = 4(3) + 1 = 13 \)
\( T_4 = 4(4) + 1 = 17 \)
\( T_5 = 4(5) + 1 = 21 \)
\( T_6 = 4(6) + 1 = 25 \)
Thus, the first six terms of the A.P. are 5, 9, 13, 17, 21, and 25.

(ii) Given that the first term \( a = 98 \) and the common difference \( d = -3 \).
The formula for the nth term of an A.P. is \( T_n = a + (n - 1)d \).
So, we find the general term \( T_n \):
\( T_n = 98 + (n - 1)(-3) \)
\( T_n = 98 - 3n + 3 \)
\( T_n = 101 - 3n \)
Now, we find the first six terms:
\( T_1 = 101 - 3(1) = 98 \)
\( T_2 = 101 - 3(2) = 95 \)
\( T_3 = 101 - 3(3) = 92 \)
\( T_4 = 101 - 3(4) = 89 \)
\( T_5 = 101 - 3(5) = 86 \)
\( T_6 = 101 - 3(6) = 83 \)
Hence, the first six terms of the A.P. are 98, 95, 92, 89, 86, and 83.

(iii) Given that the first term \( a = 7\frac { 1 }{ 2 } = \frac { 15 }{ 2 } \) and the common difference \( d = 1\frac { 1 }{ 2 } = \frac { 3 }{ 2 } \).
The formula for the nth term of an A.P. is \( T_n = a + (n - 1)d \).
So, we find the general term \( T_n \):
\( T_n = \frac { 15 }{ 2 } + (n - 1)\frac { 3 }{ 2 } \)
\( T_n = \frac { 15 + 3n - 3 }{ 2 } \)
\( T_n = \frac { 3n + 12 }{ 2 } \)
Now, we find the first six terms:
\( T_1 = \frac { 3(1) + 12 }{ 2 } = \frac { 15 }{ 2 } = 7\frac { 1 }{ 2 } \)
\( T_2 = \frac { 3(2) + 12 }{ 2 } = \frac { 18 }{ 2 } = 9 \)
\( T_3 = \frac { 3(3) + 12 }{ 2 } = \frac { 21 }{ 2 } = 10\frac { 1 }{ 2 } \)
\( T_4 = \frac { 3(4) + 12 }{ 2 } = \frac { 24 }{ 2 } = 12 \)
\( T_5 = \frac { 3(5) + 12 }{ 2 } = \frac { 27 }{ 2 } = 13\frac { 1 }{ 2 } \)
\( T_6 = \frac { 3(6) + 12 }{ 2 } = \frac { 30 }{ 2 } = 15 \)
Thus, the first six terms of the A.P. are \( 7\frac { 1 }{ 2 }, 9, 10\frac { 1 }{ 2 }, 12, 13\frac { 1 }{ 2 } \) and 15.
In simple words: To find the terms of an A.P., we start with the first term and keep adding the common difference repeatedly. For instance, the second term is the first term plus the common difference, the third term is the second term plus the common difference, and so on.

🎯 Exam Tip: Always write down the formula for the nth term \( T_n = a + (n-1)d \) at the start of your solution. This shows the examiner you know the basic principle and helps prevent errors in calculation.

Question 2. Write the 5th and 8th terms of an A.P. whose 10th term is 43 and the common difference is 4.
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the given A.P.
We are given that the common difference \( d = 4 \).
We are also given that the 10th term \( T_{10} = 43 \).
Using the formula for the nth term of an A.P., \( T_n = a + (n - 1)d \):
For the 10th term:
\( T_{10} = a + (10 - 1)d = a + 9d \)
So, we have the equation:
\( a + 9d = 43 \)
Substitute the value of \( d = 4 \) into the equation:
\( a + 9(4) = 43 \)
\( a + 36 = 43 \)
\( a = 43 - 36 \)
\( a = 7 \)
Now we have the first term \( a = 7 \) and the common difference \( d = 4 \).
We need to find the 5th term \( T_5 \):
\( T_5 = a + (5 - 1)d = a + 4d \)
\( T_5 = 7 + 4(4) \)
\( T_5 = 7 + 16 \)
\( T_5 = 23 \)
Next, we need to find the 8th term \( T_8 \):
\( T_8 = a + (8 - 1)d = a + 7d \)
\( T_8 = 7 + 7(4) \)
\( T_8 = 7 + 28 \)
\( T_8 = 35 \)
So, the 5th term is 23 and the 8th term is 35.
In simple words: We first used the given 10th term and common difference to find the starting number of the A.P. Once we had that, we could easily calculate the 5th and 8th terms by adding the common difference the right number of times.

🎯 Exam Tip: When given a term and the common difference, always calculate the first term 'a' first. This is a crucial step that simplifies finding any other term in the series.

Question 3. In each of the following find the terms required.
(a) The seventh term of 2, 7, 12, ......
(b) The fifth term of 21, 28, 35, ......
(c) The eighteenth term of 9, 5, 1, ......
Answer:
(a) Given the sequence: 2, 7, 12, ...
First term \( a = 2 \).
To find the common difference \( d \), subtract a term from its next term:
\( T_2 - T_1 = 7 - 2 = 5 \)
\( T_3 - T_2 = 12 - 7 = 5 \)
So, the common difference \( d = 5 \).
We need to find the seventh term \( T_7 \). Using the formula \( T_n = a + (n - 1)d \):
\( T_7 = a + (7 - 1)d = a + 6d \)
\( T_7 = 2 + 6(5) \)
\( T_7 = 2 + 30 \)
\( T_7 = 32 \)
The seventh term is 32.

(b) Given the sequence: 21, 28, 35, ...
First term \( a = 21 \).
To find the common difference \( d \):
\( T_2 - T_1 = 28 - 21 = 7 \)
\( T_3 - T_2 = 35 - 28 = 7 \)
So, the common difference \( d = 7 \).
We need to find the fifth term \( T_5 \). Using the formula \( T_n = a + (n - 1)d \):
\( T_5 = a + (5 - 1)d = a + 4d \)
\( T_5 = 21 + 4(7) \)
\( T_5 = 21 + 28 \)
\( T_5 = 49 \)
The fifth term is 49.

(c) Given the sequence: 9, 5, 1, ...
First term \( a = 9 \).
To find the common difference \( d \):
\( T_2 - T_1 = 5 - 9 = -4 \)
\( T_3 - T_2 = 1 - 5 = -4 \)
So, the common difference \( d = -4 \).
We need to find the eighteenth term \( T_{18} \). Using the formula \( T_n = a + (n - 1)d \):
\( T_{18} = a + (18 - 1)d = a + 17d \)
\( T_{18} = 9 + 17(-4) \)
\( T_{18} = 9 - 68 \)
\( T_{18} = -59 \)
The eighteenth term is -59.
In simple words: For each list of numbers, we first figure out the starting number and the pattern (how much is added or subtracted each time). Then, using a simple rule, we can quickly find any term in the list, no matter how far down it is.

🎯 Exam Tip: Always double-check the common difference, especially when it's negative. A small calculation error there will make all subsequent terms incorrect.

Question 4. In each of the following, the nth term is given. Find the terms required:
(a) \( 4n - 2 \), find the 11th term.
(b) \( 6n + 5 \), find the 11th term.
(c) \( 101 - 3n \), find the 11th term.
Answer:
(a) The nth term is given by \( T_n = 4n - 2 \).
To find the 11th term, we substitute \( n = 11 \) into the expression:
\( T_{11} = 4(11) - 2 \)
\( T_{11} = 44 - 2 \)
\( T_{11} = 42 \)
The 11th term is 42.

(b) The nth term is given by \( T_n = 6n + 5 \).
To find the 11th term, we substitute \( n = 11 \) into the expression:
\( T_{11} = 6(11) + 5 \)
\( T_{11} = 66 + 5 \)
\( T_{11} = 71 \)
The 11th term is 71.

(c) The nth term is given by \( T_n = 101 - 3n \).
To find the 11th term, we substitute \( n = 11 \) into the expression:
\( T_{11} = 101 - 3(11) \)
\( T_{11} = 101 - 33 \)
\( T_{11} = 68 \)
The 11th term is 68.
In simple words: When you have a rule that tells you how to find any term in a sequence (like \( T_n = 4n - 2 \)), to find a specific term like the 11th, you just replace 'n' with 11 in that rule and do the calculation.

🎯 Exam Tip: This type of question is straightforward. The most common mistake is a simple arithmetic error, so always re-check your calculations, especially with multiplication and subtraction.

Question 5. The 5th term of an A.P. is 11 and the 9th term is 7. Find the 16th term.
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the A.P.
We are given the 5th term, \( T_5 = 11 \). Using the formula \( T_n = a + (n - 1)d \):
\( T_5 = a + (5 - 1)d \)
\( a + 4d = 11 \) ...(1)

We are also given the 9th term, \( T_9 = 7 \). Using the formula \( T_n = a + (n - 1)d \):
\( T_9 = a + (9 - 1)d \)
\( a + 8d = 7 \) ...(2)

Now we have a system of two linear equations. Subtract equation (1) from equation (2) to find \( d \):
\( (a + 8d) - (a + 4d) = 7 - 11 \)
\( a + 8d - a - 4d = -4 \)
\( 4d = -4 \)
\( d = -1 \)

Now substitute the value of \( d = -1 \) into equation (1) to find \( a \):
\( a + 4(-1) = 11 \)
\( a - 4 = 11 \)
\( a = 11 + 4 \)
\( a = 15 \)

So, the first term is \( a = 15 \) and the common difference is \( d = -1 \).
We need to find the 16th term, \( T_{16} \). Using the formula \( T_n = a + (n - 1)d \):
\( T_{16} = a + (16 - 1)d \)
\( T_{16} = a + 15d \)
\( T_{16} = 15 + 15(-1) \)
\( T_{16} = 15 - 15 \)
\( T_{16} = 0 \)
The 16th term of the A.P. is 0.
In simple words: We used the information about two different terms in the sequence to first find the starting number and the step size. Once those were known, finding any other term, like the 16th term, was a straightforward calculation.

🎯 Exam Tip: When solving for 'a' and 'd' from two given terms, always write down your two equations clearly. Subtracting them is usually the easiest way to find 'd' first, then substitute back to find 'a'.

Question 6. Which term of the series 5, 8, 11, ....... is 320 ?
Answer:
Given the arithmetic series: 5, 8, 11, ...
First term \( a = 5 \).
To find the common difference \( d \):
\( T_2 - T_1 = 8 - 5 = 3 \)
\( T_3 - T_2 = 11 - 8 = 3 \)
So, the common difference \( d = 3 \).
We need to find which term \( n \) has a value of 320. Let \( T_n = 320 \).
Using the formula for the nth term of an A.P., \( T_n = a + (n - 1)d \):
\( 320 = 5 + (n - 1)3 \)
Subtract 5 from both sides:
\( 320 - 5 = (n - 1)3 \)
\( 315 = (n - 1)3 \)
Divide both sides by 3:
\( \frac { 315 }{ 3 } = n - 1 \)
\( 105 = n - 1 \)
Add 1 to both sides:
\( n = 105 + 1 \)
\( n = 106 \)
Therefore, the 106th term of the A.P. is 320.
In simple words: We know the first number and the step size of a number pattern. We want to find out which position in that pattern has the number 320. By using a simple rule, we can work backward to find its position.

🎯 Exam Tip: Always set up the equation \( T_n = a + (n-1)d \) carefully. The most common error is miscalculating \( n-1 \) or performing the subtraction/division incorrectly when solving for \( n \).

Question 7. The fourth term of an A.P. is ten times the first. Prove that the sixth term is four times as great as the second term.
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the A.P.
The formula for the nth term is \( T_n = a + (n - 1)d \).
According to the first condition, the fourth term (\( T_4 \)) is ten times the first term (\( T_1 \)):
\( T_4 = 10 \times T_1 \)
We know \( T_4 = a + (4 - 1)d = a + 3d \) and \( T_1 = a \).
So, substituting these into the condition:
\( a + 3d = 10a \)
Subtract \( a \) from both sides:
\( 3d = 9a \)
Divide by 3:
\( d = 3a \) ...(1)
Now, we need to prove that the sixth term (\( T_6 \)) is four times the second term (\( T_2 \)).
Let's find \( T_6 \) and \( T_2 \) in terms of \( a \) and \( d \):
\( T_6 = a + (6 - 1)d = a + 5d \)
\( T_2 = a + (2 - 1)d = a + d \)
Substitute \( d = 3a \) (from equation 1) into the expressions for \( T_6 \) and \( T_2 \):
\( T_6 = a + 5(3a) = a + 15a = 16a \)
\( T_2 = a + 3a = 4a \)
Now, let's compare \( T_6 \) and \( T_2 \):
\( T_6 = 16a \)
\( T_2 = 4a \)
We can see that \( 16a = 4 \times (4a) \).

\( \implies T_6 = 4 \times T_2 \)
Thus, it is proven that the sixth term is four times as great as the second term.
In simple words: We used the given hint about how the fourth term relates to the first term to find a link between the starting number and the step size. Then, we used this link to show that the sixth term is exactly four times bigger than the second term.

🎯 Exam Tip: For proof questions, clearly define 'a' and 'd' at the beginning. Express all terms mentioned in the problem using the \( T_n = a + (n-1)d \) formula. Substitute and simplify step-by-step to arrive at the required conclusion.

Question 8. The fourth term of an A.P. is equal to 3 times the first term, and the seventh term exceeds twice the third term by 1. Find the first term and the common difference.
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the A.P.
The formula for the nth term is \( T_n = a + (n - 1)d \).
According to the first condition: The fourth term (\( T_4 \)) is equal to 3 times the first term (\( T_1 \)).
\( T_4 = 3T_1 \)
\( a + 3d = 3a \)
Subtract \( a \) from both sides:
\( 3d = 2a \) ...(1)

According to the second condition: The seventh term (\( T_7 \)) exceeds twice the third term (\( T_3 \)) by 1.
\( T_7 = 2T_3 + 1 \)
\( a + (7 - 1)d = 2[a + (3 - 1)d] + 1 \)
\( a + 6d = 2(a + 2d) + 1 \)
\( a + 6d = 2a + 4d + 1 \)
Rearrange the terms to form an equation with \( a \) and \( d \):
\( 6d - 4d - 1 = 2a - a \)
\( 2d - 1 = a \) ...(2)

Now we have a system of two equations:
1) \( 2a = 3d \)
2) \( a = 2d - 1 \)

Substitute the expression for \( a \) from equation (2) into equation (1):
\( 2(2d - 1) = 3d \)
\( 4d - 2 = 3d \)
Subtract \( 3d \) from both sides:
\( 4d - 3d - 2 = 0 \)
\( d - 2 = 0 \)
\( d = 2 \)

Now substitute the value of \( d = 2 \) into equation (2) to find \( a \):
\( a = 2(2) - 1 \)
\( a = 4 - 1 \)
\( a = 3 \)
So, the first term is 3 and the common difference is 2.
In simple words: We used two pieces of information about the number pattern to create two separate math problems. By solving these two problems together, we found both the starting number and the constant step size of the pattern.

🎯 Exam Tip: Break down complex word problems into separate conditions. Formulate an algebraic equation for each condition, then solve the system of equations. Clearly label your equations (1), (2) for better clarity.

Question 9. Which term of the progression 19, \( 18\frac { 1 }{ 5 } \), \( 17\frac { 2 }{ 5 } \), ..... is the first negative term ?
Answer:
Given the progression: 19, \( 18\frac { 1 }{ 5 } \), \( 17\frac { 2 }{ 5 } \), ...
First term \( a = 19 \).
Let's convert the mixed fractions to improper fractions:
\( 18\frac { 1 }{ 5 } = \frac { (18 \times 5) + 1 }{ 5 } = \frac { 90 + 1 }{ 5 } = \frac { 91 }{ 5 } \)
\( 17\frac { 2 }{ 5 } = \frac { (17 \times 5) + 2 }{ 5 } = \frac { 85 + 2 }{ 5 } = \frac { 87 }{ 5 } \)

The sequence is: 19, \( \frac { 91 }{ 5 } \), \( \frac { 87 }{ 5 } \), ...
To find the common difference \( d \):
\( d = T_2 - T_1 = \frac { 91 }{ 5 } - 19 \)
\( d = \frac { 91 }{ 5 } - \frac { 19 \times 5 }{ 5 } = \frac { 91 - 95 }{ 5 } = \frac { -4 }{ 5 } \)

We need to find the first term \( T_n \) that is negative, which means \( T_n < 0 \).
Using the formula \( T_n = a + (n - 1)d \):
\( 19 + (n - 1)\left(\frac { -4 }{ 5 }\right) < 0 \)
Multiply the entire inequality by 5 to remove the fraction and simplify:
\( 5 \times 19 + 5 \times (n - 1)\left(\frac { -4 }{ 5 }\right) < 5 \times 0 \)
\( 95 - 4(n - 1) < 0 \)
\( 95 - 4n + 4 < 0 \)
\( 99 - 4n < 0 \)
Add \( 4n \) to both sides:
\( 99 < 4n \)
Divide by 4:
\( \frac { 99 }{ 4 } < n \)
\( 24.75 < n \)
Since \( n \) must be a natural number (it represents the term number), the smallest integer value for \( n \) that is greater than 24.75 is 25.
Therefore, the 25th term is the first negative term.
In simple words: This number pattern starts big and gets smaller each time. We wanted to find when the numbers would first go below zero. By setting up a rule and solving it, we found that the 25th number in the pattern would be the first one to become negative.

🎯 Exam Tip: When dealing with inequalities to find 'n', remember that 'n' must be a positive integer. If you multiply or divide an inequality by a negative number, you must reverse the inequality sign. Always consider the smallest integer greater than your calculated 'n' if 'n' is not an integer.

Question 10.
(i) Find the value of k so that \( 8k + 4 \), \( 6k - 2 \), and \( 2k + 1 \) will form an A.P.
(ii) Find a, b such that 7.2, a, b, 3 are in A.P. (SC)
Answer:
(i) If three terms, say x, y, z, are in an A.P., then the middle term is the average of the other two, i.e., \( 2y = x + z \).
Given terms are \( 8k + 4 \), \( 6k - 2 \), and \( 2k + 1 \).
So, we can write:
\( 2(6k - 2) = (8k + 4) + (2k + 1) \)
Now, we solve for \( k \):
\( 12k - 4 = 10k + 5 \)
Subtract \( 10k \) from both sides:
\( 12k - 10k - 4 = 5 \)
\( 2k - 4 = 5 \)
Add 4 to both sides:
\( 2k = 5 + 4 \)
\( 2k = 9 \)
\( k = \frac { 9 }{ 2 } \)
Thus, the value of \( k \) is \( \frac { 9 }{ 2 } \).

(ii) Given terms in A.P. are 7.2, a, b, 3.
Since these terms are in A.P., the common difference between consecutive terms must be the same.
\( a - 7.2 = d \) ...(1)
\( b - a = d \) ...(2)
\( 3 - b = d \) ...(3)

From (1) and (2):
\( a - 7.2 = b - a \)
\( 2a - b = 7.2 \) ...(4)

From (2) and (3):
\( b - a = 3 - b \)
\( 2b - a = 3 \) ...(5)

Now we have a system of two linear equations:
4) \( 2a - b = 7.2 \)
5) \( -a + 2b = 3 \)

Multiply equation (5) by 2:
\( 2(-a + 2b) = 2(3) \)
\( -2a + 4b = 6 \) ...(6)

Add equation (4) and equation (6):
\( (2a - b) + (-2a + 4b) = 7.2 + 6 \)
\( 3b = 13.2 \)
\( b = \frac { 13.2 }{ 3 } \)
\( b = 4.4 \)

Substitute \( b = 4.4 \) into equation (5):
\( -a + 2(4.4) = 3 \)
\( -a + 8.8 = 3 \)
\( -a = 3 - 8.8 \)
\( -a = -5.8 \)
\( a = 5.8 \)
Therefore, \( a = 5.8 \) and \( b = 4.4 \).
In simple words: For the first part, we used the rule that in an A.P., the middle number is the average of its neighbors. For the second part, we set up equations based on the idea that the difference between any two consecutive numbers in an A.P. is always the same. Solving these equations helped us find the missing numbers.

🎯 Exam Tip: Remember the property of an A.P.: if a, b, c are in A.P., then \( 2b = a + c \). This is very useful for finding unknown values in sequences. For more terms, create simultaneous equations from equal common differences.

Question 11. Determine 2nd term and rth term of an A.P. whose 6th term is 12 and 8th term is 22.
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the A.P.
The formula for the nth term is \( T_n = a + (n - 1)d \).
We are given the 6th term, \( T_6 = 12 \):
\( T_6 = a + (6 - 1)d \)
\( a + 5d = 12 \) ...(1)

We are also given the 8th term, \( T_8 = 22 \):
\( T_8 = a + (8 - 1)d \)
\( a + 7d = 22 \) ...(2)

Now, we solve this system of equations. Subtract equation (1) from equation (2):
\( (a + 7d) - (a + 5d) = 22 - 12 \)
\( a + 7d - a - 5d = 10 \)
\( 2d = 10 \)
\( d = 5 \)

Substitute the value of \( d = 5 \) into equation (1):
\( a + 5(5) = 12 \)
\( a + 25 = 12 \)
\( a = 12 - 25 \)
\( a = -13 \)

So, the first term is \( a = -13 \) and the common difference is \( d = 5 \).
Now, we need to determine the 2nd term (\( T_2 \)):
\( T_2 = a + (2 - 1)d = a + d \)
\( T_2 = -13 + 5 \)
\( T_2 = -8 \)

Next, we determine the rth term (\( T_r \)):
\( T_r = a + (r - 1)d \)
\( T_r = -13 + (r - 1)5 \)
\( T_r = -13 + 5r - 5 \)
\( T_r = 5r - 18 \)
The 2nd term is -8 and the rth term is \( 5r - 18 \).
In simple words: We used the given values of two different terms to find the starting number and the amount added each time in the sequence. Once we knew these, we could easily find the second term and also a general rule for any 'r-th' term.

🎯 Exam Tip: Always solve for 'a' and 'd' first when you have information about two terms. Then, you can easily find any other term or the general expression for the nth term. Be careful with signs when 'a' is negative.

Question 12. Prove that the product of the 2nd and 3rd terms of an A.P. exceeds the product of the 1st and 4th by twice the square of the difference between the 1st and 2nd term.
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the A.P.
The terms of the A.P. are:
First term (\( T_1 \)) = \( a \)
Second term (\( T_2 \)) = \( a + d \)
Third term (\( T_3 \)) = \( a + 2d \)
Fourth term (\( T_4 \)) = \( a + 3d \)

We need to prove that \( T_2 \cdot T_3 = T_1 \cdot T_4 + 2(T_2 - T_1)^2 \).
Let's calculate the left-hand side (LHS):
LHS = Product of the 2nd and 3rd terms = \( T_2 \cdot T_3 \)
\( = (a + d)(a + 2d) \)
\( = a^2 + 2ad + ad + 2d^2 \)
\( = a^2 + 3ad + 2d^2 \) ...(1)

Now, let's calculate the right-hand side (RHS):
RHS = Product of the 1st and 4th terms + twice the square of the difference between the 1st and 2nd term
\( = T_1 \cdot T_4 + 2(T_2 - T_1)^2 \)
First, calculate \( T_2 - T_1 \):
\( T_2 - T_1 = (a + d) - a = d \)
So, \( (T_2 - T_1)^2 = d^2 \)
Now substitute this into the RHS:
\( = a(a + 3d) + 2(d)^2 \)
\( = a^2 + 3ad + 2d^2 \) ...(2)

From (1) and (2), we can see that LHS = RHS.
\( a^2 + 3ad + 2d^2 = a^2 + 3ad + 2d^2 \)
Hence, the product of the 2nd and 3rd terms of an A.P. exceeds the product of the 1st and 4th by twice the square of the difference between the 1st and 2nd term.
In simple words: We took the first four numbers in a pattern and created an equation using them. By carefully replacing each number with its general form (like 'a' for the first number and 'd' for the step size), we showed that both sides of the equation always turn out to be the same, proving the statement true.

🎯 Exam Tip: For proof questions involving A.P. terms, always express all terms (\( T_1, T_2, T_3, T_4 \), etc.) in terms of 'a' (first term) and 'd' (common difference). Simplify both sides of the equation you need to prove separately, then show they are equal.

Question 13. The 2nd, 31st and last term of an A.P. are \( 7\frac { 3 }{ 4 } \), \( \frac { 1 }{ 2 } \) and \( -6\frac { 1 }{ 2 } \) respectively. Find the first term and the number of terms.
Answer:
Let \( a \) be the first term, \( d \) be the common difference, and \( n \) be the number of terms in the A.P.
The formula for the nth term is \( T_n = a + (n - 1)d \).
We are given the 2nd term: \( T_2 = 7\frac { 3 }{ 4 } \)
\( 7\frac { 3 }{ 4 } = \frac { (7 \times 4) + 3 }{ 4 } = \frac { 28 + 3 }{ 4 } = \frac { 31 }{ 4 } \)
So, \( a + d = \frac { 31 }{ 4 } \) ...(1)

We are given the 31st term: \( T_{31} = \frac { 1 }{ 2 } \)
So, \( a + 30d = \frac { 1 }{ 2 } \) ...(2)

Now we have a system of two linear equations. Subtract equation (1) from equation (2):
\( (a + 30d) - (a + d) = \frac { 1 }{ 2 } - \frac { 31 }{ 4 } \)
\( 29d = \frac { 2 }{ 4 } - \frac { 31 }{ 4 } \)
\( 29d = \frac { -29 }{ 4 } \)
Divide by 29:
\( d = \frac { -29 }{ 4 \times 29 } \)
\( d = -\frac { 1 }{ 4 } \)

Substitute the value of \( d = -\frac { 1 }{ 4 } \) into equation (1) to find \( a \):
\( a + \left(-\frac { 1 }{ 4 }\right) = \frac { 31 }{ 4 } \)
\( a - \frac { 1 }{ 4 } = \frac { 31 }{ 4 } \)
\( a = \frac { 31 }{ 4 } + \frac { 1 }{ 4 } \)
\( a = \frac { 32 }{ 4 } \)
\( a = 8 \)
So, the first term is \( a = 8 \).

We are given that the last term \( T_n = -6\frac { 1 }{ 2 } \)
\( -6\frac { 1 }{ 2 } = -\frac { (6 \times 2) + 1 }{ 2 } = -\frac { 12 + 1 }{ 2 } = -\frac { 13 }{ 2 } \)
Now, use the formula \( T_n = a + (n - 1)d \) to find \( n \):
\( -\frac { 13 }{ 2 } = 8 + (n - 1)\left(-\frac { 1 }{ 4 }\right) \)
Subtract 8 from both sides:
\( -\frac { 13 }{ 2 } - 8 = (n - 1)\left(-\frac { 1 }{ 4 }\right) \)
\( -\frac { 13 }{ 2 } - \frac { 16 }{ 2 } = (n - 1)\left(-\frac { 1 }{ 4 }\right) \)
\( -\frac { 29 }{ 2 } = (n - 1)\left(-\frac { 1 }{ 4 }\right) \)
Multiply both sides by -4:
\( -\frac { 29 }{ 2 } \times (-4) = n - 1 \)
\( 29 \times 2 = n - 1 \)
\( 58 = n - 1 \)
\( n = 58 + 1 \)
\( n = 59 \)
The first term is 8 and the number of terms is 59.
In simple words: We used the given second and thirty-first terms to find the starting number and the step size of the number pattern. Then, using the last term given, we calculated how many numbers are in the entire pattern.

🎯 Exam Tip: Always convert mixed fractions to improper fractions before calculations to avoid errors. When dealing with negative common differences, be extra careful with arithmetic operations, especially multiplication and division.

Question 14. If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, show that the 18th term of the A.P. is zero.
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the A.P.
The formula for the nth term is \( T_n = a + (n - 1)d \).
The 7th term is \( T_7 = a + (7 - 1)d = a + 6d \).
The 11th term is \( T_{11} = a + (11 - 1)d = a + 10d \).
We are given the condition:
7 times the 7th term is equal to 11 times the 11th term.
\( 7 \times T_7 = 11 \times T_{11} \)
Substitute the expressions for \( T_7 \) and \( T_{11} \):
\( 7(a + 6d) = 11(a + 10d) \)
Distribute the numbers:
\( 7a + 42d = 11a + 110d \)
Rearrange the terms to one side of the equation:
\( 0 = 11a - 7a + 110d - 42d \)
\( 0 = 4a + 68d \)
Factor out 4 from the right side:
\( 0 = 4(a + 17d) \)
Divide by 4:
\( 0 = a + 17d \)
Now, let's consider the 18th term, \( T_{18} \):
\( T_{18} = a + (18 - 1)d = a + 17d \)
Since we found that \( a + 17d = 0 \), this means:
\( T_{18} = 0 \)
Thus, it is shown that the 18th term of the A.P. is zero.
In simple words: We used the given relationship between the 7th and 11th terms to create an equation. By solving and simplifying this equation, we discovered that the sum of the first term and seventeen times the common difference is zero, which is exactly how the 18th term is calculated.

🎯 Exam Tip: This type of proof is common. Clearly write out the expressions for \( T_n \) for the given terms. Carefully expand and simplify the given condition to see if it directly leads to the expression for the term you need to prove. Watch out for algebraic mistakes.

Question 15. If \( k + 2 \), \( 4k - 6 \) and \( 3k - 2 \) are three consecutive terms of an A.P., find the value of k.
Answer:
If three terms, \( x, y, z \), are in an A.P., then the common difference between consecutive terms must be the same. This means \( y - x = z - y \), which simplifies to \( 2y = x + z \).
Given the three consecutive terms of an A.P.:
First term \( x = k + 2 \)
Second term \( y = 4k - 6 \)
Third term \( z = 3k - 2 \)
Using the property \( 2y = x + z \):
\( 2(4k - 6) = (k + 2) + (3k - 2) \)
Now, we solve for \( k \):
On the left side:
\( 8k - 12 \)
On the right side:
\( k + 2 + 3k - 2 = 4k \)
So, the equation becomes:
\( 8k - 12 = 4k \)
Subtract \( 4k \) from both sides:
\( 8k - 4k - 12 = 0 \)
\( 4k - 12 = 0 \)
Add 12 to both sides:
\( 4k = 12 \)
Divide by 4:
\( k = \frac { 12 }{ 4 } \)
\( k = 3 \)
The value of \( k \) is 3.
In simple words: In any list of numbers that follows a pattern where you add the same amount each time, the middle number is always exactly halfway between the numbers before and after it. We used this simple rule to find the unknown value 'k'.

🎯 Exam Tip: The key property for three consecutive terms a, b, c in an A.P. is that \( 2b = a + c \). Applying this rule correctly is crucial for solving such problems. Be careful with distributing numbers and combining like terms.

Question 16. The pth term of an A.P. is q and the qth term is p, show that the mth term is \( p + q - m \).
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the A.P.
The formula for the nth term is \( T_n = a + (n - 1)d \).
We are given that the pth term is q:
\( T_p = a + (p - 1)d = q \) ...(1)
We are also given that the qth term is p:
\( T_q = a + (q - 1)d = p \) ...(2)

To find \( d \), subtract equation (2) from equation (1):
\( [a + (p - 1)d] - [a + (q - 1)d] = q - p \)
\( a + pd - d - a - qd + d = q - p \)
\( pd - qd = q - p \)
\( d(p - q) = q - p \)
\( d(p - q) = -(p - q) \)
If \( p \neq q \), we can divide by \( (p - q) \):
\( d = -1 \)

Now, substitute \( d = -1 \) into equation (1) to find \( a \):
\( a + (p - 1)(-1) = q \)
\( a - p + 1 = q \)
\( a = p + q - 1 \)

Now we need to find the mth term, \( T_m \). Using the formula \( T_m = a + (m - 1)d \):
Substitute the values of \( a \) and \( d \) we found:
\( T_m = (p + q - 1) + (m - 1)(-1) \)
\( T_m = p + q - 1 - m + 1 \)
\( T_m = p + q - m \)
Thus, it is shown that the mth term is \( p + q - m \).
In simple words: We used the given facts about the 'p-th' and 'q-th' terms to find the starting number and the step size of the pattern. Once we had those, we put them into the rule for finding any 'm-th' term, and it simplified to exactly what we needed to prove. This problem assumes \(p \neq q\), otherwise there would be an infinite number of solutions or no solution.

🎯 Exam Tip: This is a standard problem type for A.P. proofs. Always solve for 'd' first by subtracting the two equations, then substitute 'd' back into one of the original equations to solve for 'a'. Finally, substitute both 'a' and 'd' into the expression for the term you need to prove.

Question 17. Let \( T_r \) be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, \( T_m = \frac { 1 }{ n } \), \( T_n = \frac { 1 }{ m } \), then \( (a - d) \) equals
(a) \( \frac { 1 }{ mn } \)
(b) 1
(c) 0
(d) \( \frac { 1 }{ m } + \frac { 1 }{ n } \)
Answer: (c) 0
Let \( a \) be the first term and \( d \) be the common difference of the A.P.
The formula for the nth term is \( T_k = a + (k - 1)d \).
We are given:
\( T_m = \frac { 1 }{ n } \implies a + (m - 1)d = \frac { 1 }{ n } \) ...(1)
\( T_n = \frac { 1 }{ m } \implies a + (n - 1)d = \frac { 1 }{ m } \) ...(2)

Subtract equation (2) from equation (1) to find \( d \):
\( [a + (m - 1)d] - [a + (n - 1)d] = \frac { 1 }{ n } - \frac { 1 }{ m } \)
\( a + md - d - a - nd + d = \frac { m - n }{ mn } \)
\( md - nd = \frac { m - n }{ mn } \)
\( d(m - n) = \frac { m - n }{ mn } \)
If \( m \neq n \), divide both sides by \( (m - n) \):
\( d = \frac { 1 }{ mn } \)

Now substitute the value of \( d = \frac { 1 }{ mn } \) into equation (1) to find \( a \):
\( a + (m - 1)\left(\frac { 1 }{ mn }\right) = \frac { 1 }{ n } \)
\( a + \frac { m }{ mn } - \frac { 1 }{ mn } = \frac { 1 }{ n } \)
\( a + \frac { 1 }{ n } - \frac { 1 }{ mn } = \frac { 1 }{ n } \)
Subtract \( \frac { 1 }{ n } \) from both sides:
\( a - \frac { 1 }{ mn } = 0 \)
\( a = \frac { 1 }{ mn } \)

Now we need to find the value of \( (a - d) \):
\( a - d = \frac { 1 }{ mn } - \frac { 1 }{ mn } \)
\( a - d = 0 \)
In simple words: We used the given rules for the 'm-th' and 'n-th' terms to find both the starting number ('a') and the step size ('d') of the pattern. It turned out that 'a' and 'd' were exactly the same value, so when we subtracted them, the answer was zero.

🎯 Exam Tip: This is a classic "common terms" problem in A.P. Always remember that for \( T_m = 1/n \) and \( T_n = 1/m \), the common difference 'd' will be \( 1/(mn) \) and the first term 'a' will also be \( 1/(mn) \). This can save time in MCQs.

Question 18. Given that the \( (p + 1) \)th term of an A.P. is twice the \( (q + 1) \)th term, prove that the \( (3p + 1) \)th term is twice the \( (p + q + 1) \)th term.
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the A.P.
The formula for the nth term is \( T_n = a + (n - 1)d \).
Given that the \( (p + 1) \)th term is twice the \( (q + 1) \)th term:
\( T_{p+1} = 2 \times T_{q+1} \)
Substitute the formula for \( T_n \):
\( a + ((p + 1) - 1)d = 2[a + ((q + 1) - 1)d] \)
\( a + pd = 2(a + qd) \)
\( a + pd = 2a + 2qd \)
Rearrange to express \( a \) in terms of \( d \):
\( pd - 2qd = 2a - a \)
\( a = (p - 2q)d \) ...(1)

Now, we need to prove that \( T_{3p+1} = 2 \times T_{p+q+1} \).
Let's find the expression for \( T_{3p+1} \):
\( T_{3p+1} = a + ((3p + 1) - 1)d \)
\( T_{3p+1} = a + 3pd \)
Substitute \( a = (p - 2q)d \) from equation (1):
\( T_{3p+1} = (p - 2q)d + 3pd \)
\( T_{3p+1} = pd - 2qd + 3pd \)
\( T_{3p+1} = (4p - 2q)d \)
\( T_{3p+1} = 2(2p - q)d \) ...(2)

Next, let's find the expression for \( T_{p+q+1} \):
\( T_{p+q+1} = a + ((p + q + 1) - 1)d \)
\( T_{p+q+1} = a + (p + q)d \)
Substitute \( a = (p - 2q)d \) from equation (1):
\( T_{p+q+1} = (p - 2q)d + (p + q)d \)
\( T_{p+q+1} = pd - 2qd + pd + qd \)
\( T_{p+q+1} = (2p - q)d \) ...(3)

From equation (2) and (3), we can see the relationship:
\( T_{3p+1} = 2(2p - q)d \)
\( T_{p+q+1} = (2p - q)d \)

\( \implies T_{3p+1} = 2 \times T_{p+q+1} \)
Hence, it is proven that the \( (3p + 1) \)th term is twice the \( (p + q + 1) \)th term.
In simple words: We used the first hint, which linked one specific term to another, to find a simple connection between the starting number of the pattern and its step size. Then, we used this connection to show that a third specific term is always exactly twice as big as a fourth specific term.

🎯 Exam Tip: When proving relationships between terms in an A.P., always simplify the initial given condition to express 'a' in terms of 'd' (or vice versa). This simplified relationship will be key for substituting into the expressions for the terms you need to prove, leading to a much clearer solution.