Samacheer Kalvi Class 10 Maths Solutions Chapter 2 Numbers and Sequences Exercise 2.6

Question 1. Find the sum of the following
(i) 3, 7, 11,... up to 40 terms.
(ii) 102, 97, 92,... up to 27 terms.
(iii) 6 + 13 + 20 + .... + 97
Answer:
(i) For the series 3, 7, 11,... up to 40 terms:
The first term \( a = 3 \).
The common difference \( d = 7 - 3 = 4 \).
The number of terms \( n = 40 \).
The sum of an arithmetic progression (AP) is given by the formula: \( S_n = \frac { n }{ 2 } [2a + (n - 1) d] \)
Substitute the values: \( S_{40} = \frac {40}{ 2 } [2(3) + (40 - 1) 4] \)
\( S_{40} = 20 [6 + 39 \times 4] \)
\( S_{40} = 20 [6 + 156] \)
\( S_{40} = 20 \times 162 \)
\( S_{40} = 3240 \)

(ii) For the series 102, 97, 92,... up to 27 terms:
The first term \( a = 102 \).
The common difference \( d = 97 - 102 = -5 \).
The number of terms \( n = 27 \).
Using the sum formula: \( S_n = \frac { n }{ 2 } [2a + (n - 1)d] \)
Substitute the values: \( S_{27} = \frac {27 }{ 2 } [2(102) + (27 - 1)(-5)] \)
\( S_{27} = \frac { 27 }{ 2 } [204 + 26(-5)] \)
\( S_{27} = \frac { 27 }{ 2 } [204 - 130] \)
\( S_{27} = \frac { 27 }{ 2 } \times 74 \)
\( S_{27} = 27 \times 37 \)
\( S_{27} = 999 \)

(iii) For the series 6 + 13 + 20 + ... + 97:
The first term \( a = 6 \).
The common difference \( d = 13 - 6 = 7 \).
The last term \( l = 97 \).
First, find the number of terms \( n \) using the formula: \( n = \frac { l-a }{ d } + 1 \)
Substitute the values: \( n = \frac { 97-6 }{ 7 } + 1 \)
\( n = \frac { 91 }{ 7 } + 1 \)
\( n = 13 + 1 \)
\( n = 14 \)
Now, find the sum using the formula: \( S_n = \frac { n }{ 2 } (a + l) \)
Substitute the values: \( S_{14} = \frac { 14 }{ 2 } (6 + 97) \)
\( S_{14} = 7 \times 103 \)
\( S_{14} = 721 \)
In simple words: For each series, we find the first term, the common difference, and the number of terms. Then, we use the correct formula for the sum of an arithmetic progression to calculate the total. Make sure to use the last term formula to find 'n' if it's not directly given.

🎯 Exam Tip: Carefully identify if the given series is an Arithmetic Progression (AP) and then use the correct formula for sum (either with 'd' or with 'l') depending on the information provided.

Question 2. How many consecutive odd integers beginning with 5 will sum to 480?
Answer:
Let the first term be \( a = 5 \).
Since these are consecutive odd integers, the common difference \( d = 2 \) (e.g., 5, 7, 9,...).
The sum of these integers is given as \( S_n = 480 \).
The formula for the sum of an AP is \( S_n = \frac { n }{ 2 } [2a + (n-1) d] \).
Substitute the known values into the formula:
\( 480 = \frac { n }{ 2 } [2(5) + (n-1) 2] \)
Multiply both sides by 2:
\( 960 = n [10 + 2n - 2] \)
\( 960 = n [8 + 2n] \)
\( 960 = 8n + 2n^2 \)
Rearrange the terms to form a quadratic equation:
\( 2n^2 + 8n - 960 = 0 \)
Divide the entire equation by 2:
\( n^2 + 4n - 480 = 0 \)
Factorize the quadratic equation. We need two numbers that multiply to -480 and add up to 4. These numbers are 24 and -20.
\( n^2 + 24n - 20n - 480 = 0 \)
\( n(n + 24) - 20(n + 24) = 0 \)
\( (n - 20)(n + 24) = 0 \)
This gives two possible values for \( n \):
\( n - 20 = 0 \implies n = 20 \)
\( n + 24 = 0 \implies n = -24 \)
Since the number of terms cannot be negative, we discard \( n = -24 \).
Thus, the number of consecutive odd integers is 20.
In simple words: We are looking for how many odd numbers, starting from 5, will add up to 480. We use the sum formula for a list of numbers. After putting in the numbers we know, we get an equation. We solve this equation to find the number of terms, which turns out to be 20. We ignore the negative answer because you cannot have a negative count of numbers.

🎯 Exam Tip: When solving for the number of terms (\( n \)), always remember that \( n \) must be a positive integer. Discard any negative or fractional solutions.

Question 3. Find the sum of first 28 terms of an A.P. whose \( n^{th} \) term is \( 4n – 3 \).
Answer:
We are given that the \( n^{th} \) term of the A.P. is \( t_n = 4n - 3 \).
We need to find the sum of the first 28 terms, so \( n = 28 \).
First, find the first term (\( a \)) by setting \( n = 1 \):
\( t_1 = 4(1) - 3 = 4 - 3 = 1 \)
So, the first term \( a = 1 \).
Next, find the second term (\( t_2 \)) to calculate the common difference:
\( t_2 = 4(2) - 3 = 8 - 3 = 5 \)
The common difference \( d = t_2 - t_1 = 5 - 1 = 4 \).
Now, we can find the \( 28^{th} \) term (\( t_{28} \)) which is the last term (\( l \)):
\( t_{28} = 4(28) - 3 = 112 - 3 = 109 \)
So, the last term \( l = 109 \).
The sum of the first \( n \) terms of an A.P. can be found using the formula: \( S_n = \frac{n}{2} (a + l) \)
Substitute the values: \( S_{28} = \frac{28}{2} (1 + 109) \)
\( S_{28} = 14 (110) \)
\( S_{28} = 1540 \)
Alternatively, using the formula \( S_n = \frac{n}{2} (2a + (n-1)d) \):
\( S_{28} = \frac{28}{2} [2(1) + (28-1)4] \)
\( S_{28} = 14 [2 + 27 \times 4] \)
\( S_{28} = 14 [2 + 108] \)
\( S_{28} = 14 [110] \)
\( S_{28} = 1540 \)
In simple words: First, we use the formula for the \( n^{th} \) term to find the very first number in the list and the common difference between numbers. Then, we find the 28th number, which is the last one we need for the sum. Finally, we use the sum formula for a list of numbers to add up all 28 terms.

🎯 Exam Tip: When given the \( n^{th} \) term formula, always find the first term (\( a \)) and common difference (\( d \)) first. You can use either sum formula, but \( S_n = \frac{n}{2} (a+l) \) is often quicker if \( l \) is easily found.

Question 4. The sum of first \( n \) terms of a certain series is given as \( 2n^2 – 3n \). Show that the series is an A.P.
Answer:
We are given the sum of the first \( n \) terms, \( S_n = 2n^2 - 3n \).
To show that the series is an A.P., we need to find its \( n^{th} \) term, \( t_n \), and then verify that its common difference is constant.
The \( n^{th} \) term can be found using the formula: \( t_n = S_n - S_{n-1} \).
First, find \( S_{n-1} \) by replacing \( n \) with \( n-1 \) in the \( S_n \) formula:
\( S_{n-1} = 2(n-1)^2 - 3(n-1) \)
\( S_{n-1} = 2(n^2 - 2n + 1) - 3n + 3 \)
\( S_{n-1} = 2n^2 - 4n + 2 - 3n + 3 \)
\( S_{n-1} = 2n^2 - 7n + 5 \)
Now, calculate \( t_n \):
\( t_n = S_n - S_{n-1} \)
\( t_n = (2n^2 - 3n) - (2n^2 - 7n + 5) \)
\( t_n = 2n^2 - 3n - 2n^2 + 7n - 5 \)
\( t_n = 4n - 5 \)
So, the \( n^{th} \) term is \( t_n = 4n - 5 \).
Now, let's find the first few terms of the series:
For \( n=1 \): \( t_1 = 4(1) - 5 = 4 - 5 = -1 \)
For \( n=2 \): \( t_2 = 4(2) - 5 = 8 - 5 = 3 \)
For \( n=3 \): \( t_3 = 4(3) - 5 = 12 - 5 = 7 \)
For \( n=4 \): \( t_4 = 4(4) - 5 = 16 - 5 = 11 \)
The series is -1, 3, 7, 11,....
Let's check the common difference (\( d \)):
\( d = t_2 - t_1 = 3 - (-1) = 3 + 1 = 4 \)
\( d = t_3 - t_2 = 7 - 3 = 4 \)
\( d = t_4 - t_3 = 11 - 7 = 4 \)
Since the common difference is constant (which is 4), the series is an Arithmetic Progression.
In simple words: To prove a series is an AP when given its sum formula, we first find the formula for its individual terms. We do this by subtracting the sum of \( n-1 \) terms from the sum of \( n \) terms. This gives us \( t_n = 4n-5 \). Since this formula for \( t_n \) is a linear expression in \( n \), it means the difference between consecutive terms will always be the same, which is the definition of an Arithmetic Progression. We also showed this by calculating the first few terms and checking their difference.

🎯 Exam Tip: Remember the key relationship \( t_n = S_n - S_{n-1} \). If \( t_n \) turns out to be a linear expression of \( n \) (like \( An + B \)), then the series is an A.P., and its common difference is \( A \).

Question 5. The 104th term and 4th term of an A.P are 125 and 0. Find the sum of first 35 terms?
Answer:
Let the first term of the A.P. be \( a \) and the common difference be \( d \).
The formula for the \( n^{th} \) term of an A.P. is \( t_n = a + (n - 1)d \).
We are given:
\( t_{104} = 125 \)
So, \( a + (104 - 1)d = 125 \implies a + 103d = 125 \) (Equation 1)
We are also given:
\( t_4 = 0 \)
So, \( a + (4 - 1)d = 0 \implies a + 3d = 0 \) (Equation 2)
Now, we solve these two linear equations simultaneously to find \( a \) and \( d \).
Subtract Equation 2 from Equation 1:
\( (a + 103d) - (a + 3d) = 125 - 0 \)
\( 100d = 125 \)
\( d = \frac{125}{100} = \frac{5}{4} \)
Substitute the value of \( d = \frac{5}{4} \) into Equation 2:
\( a + 3 \left( \frac{5}{4} \right) = 0 \)
\( a + \frac{15}{4} = 0 \)
\( a = -\frac{15}{4} \)
Now we have the first term \( a = -\frac{15}{4} \) and the common difference \( d = \frac{5}{4} \).
We need to find the sum of the first 35 terms (\( S_{35} \)).
The formula for the sum of \( n \) terms of an A.P. is \( S_n = \frac{n}{2} [2a + (n-1)d] \).
Substitute \( n = 35 \), \( a = -\frac{15}{4} \), and \( d = \frac{5}{4} \):
\( S_{35} = \frac{35}{2} \left[ 2 \left(-\frac{15}{4}\right) + (35-1)\left(\frac{5}{4}\right) \right] \)
\( S_{35} = \frac{35}{2} \left[ -\frac{15}{2} + 34\left(\frac{5}{4}\right) \right] \)
\( S_{35} = \frac{35}{2} \left[ -\frac{15}{2} + \frac{170}{4} \right] \)
\( S_{35} = \frac{35}{2} \left[ -\frac{15}{2} + \frac{85}{2} \right] \)
\( S_{35} = \frac{35}{2} \left[ \frac{-15 + 85}{2} \right] \)
\( S_{35} = \frac{35}{2} \left[ \frac{70}{2} \right] \)
\( S_{35} = \frac{35}{2} \times 35 \)
\( S_{35} = \frac{1225}{2} \)
\( S_{35} = 612.5 \)
The sum of the first 35 terms is 612.5.
In simple words: We are given information about two specific terms in a list of numbers (an AP). We use these two pieces of information to create two simple equations. By solving these equations, we find the starting number and the pattern of increase (or decrease) for the list. Once we have these, we use the sum formula to add up the first 35 numbers in that specific list.

🎯 Exam Tip: When given two terms of an A.P., always form two equations using \( t_n = a + (n-1)d \) and solve them simultaneously to find \( a \) and \( d \) first. This is a common setup for such problems.

Question 6. Find the sum of all odd positive integers less than 450.
Answer:
The odd positive integers less than 450 form an A.P.: 1, 3, 5, ..., 449.
Here, the first term \( a = 1 \).
The common difference \( d = 3 - 1 = 2 \).
The last term \( l = 449 \).
First, find the number of terms \( n \) using the formula: \( n = \frac { l-a }{ d } + 1 \)
\( n = \frac { 449-1 }{ 2 } + 1 \)
\( n = \frac { 448 }{ 2 } + 1 \)
\( n = 224 + 1 \)
\( n = 225 \)
Now, find the sum of these \( n \) terms using the formula: \( S_n = \frac { n }{ 2 } (a + l) \)
\( S_{225} = \frac { 225 }{ 2 } (1 + 449) \)
\( S_{225} = \frac { 225 }{ 2 } (450) \)
\( S_{225} = 225 \times 225 \)
\( S_{225} = 50625 \)
So, the sum of all odd positive integers less than 450 is 50625.
Alternatively, for the sum of the first \( n \) odd positive integers, the formula is \( S_n = n^2 \).
Since there are 225 odd positive integers less than 450, \( n = 225 \).
So, \( S_{225} = 225^2 = 50625 \).
In simple words: We need to add up all the odd numbers starting from 1 until just before 450. We find out how many such numbers there are. Then, we use a simple sum formula for an arithmetic progression. A useful trick for adding the first 'n' odd numbers is just to square 'n'. Both methods give us the same total.

🎯 Exam Tip: Recognize that the sum of the first \( n \) odd positive integers is always \( n^2 \). This shortcut can save time, but showing the A.P. method is also perfectly valid.

Question 7. Find the sum of all natural numbers between 602 and 902 which are not divisible by 4?
Answer:
To find the sum of natural numbers between 602 and 902 which are not divisible by 4, we will use the following steps:
1. Find the sum of all natural numbers between 602 and 902.
2. Find the sum of all natural numbers between 602 and 902 which are divisible by 4.
3. Subtract the second sum from the first sum.

**Step 1: Sum of all natural numbers between 602 and 902**
The natural numbers between 602 and 902 are 603, 604, ..., 901.
This is an A.P. with:
First term \( a = 603 \).
Common difference \( d = 1 \).
Last term \( l = 901 \).
First, find the number of terms \( n \): \( n = \frac { l-a }{ d } + 1 \)
\( n = \frac { 901-603 }{ 1 } + 1 \)
\( n = 298 + 1 \)
\( n = 299 \)
Now, find the sum \( S_{n1} \) using \( S_n = \frac { n }{ 2 } (a + l) \):
\( S_{n1} = \frac { 299 }{ 2 } (603 + 901) \)
\( S_{n1} = \frac { 299 }{ 2 } (1504) \)
\( S_{n1} = 299 \times 752 \)
\( S_{n1} = 224848 \)

**Step 2: Sum of all natural numbers between 602 and 902 which are divisible by 4**
The first number greater than 602 divisible by 4 is 604.
The last number less than 902 divisible by 4 is 900.
This is an A.P. with:
First term \( a = 604 \).
Common difference \( d = 4 \).
Last term \( l = 900 \).
First, find the number of terms \( n \): \( n = \frac { l-a }{ d } + 1 \)
\( n = \frac { 900-604 }{ 4 } + 1 \)
\( n = \frac { 296 }{ 4 } + 1 \)
\( n = 74 + 1 \)
\( n = 75 \)
Now, find the sum \( S_{n2} \) using \( S_n = \frac { n }{ 2 } (a + l) \):
\( S_{n2} = \frac { 75 }{ 2 } (604 + 900) \)
\( S_{n2} = \frac { 75 }{ 2 } (1504) \)
\( S_{n2} = 75 \times 752 \)
\( S_{n2} = 56400 \)

**Step 3: Sum of numbers not divisible by 4**
Sum of numbers not divisible by 4 = (Sum of all numbers) - (Sum of numbers divisible by 4)
Sum = \( S_{n1} - S_{n2} \)
Sum = \( 224848 - 56400 \)
Sum = \( 168448 \)
In simple words: To find the sum of numbers that are NOT divisible by 4, we first find the sum of ALL numbers in the given range. Then, we find the sum of only those numbers in the range that ARE divisible by 4. Finally, we subtract the second sum from the first sum to get our answer. This method works well for "not divisible by" problems.

🎯 Exam Tip: For "not divisible by" problems within a range, calculate the total sum, then the sum of terms *divisible* by the number, and subtract. Be careful with inclusive/exclusive range definitions (e.g., "between X and Y" usually means X and Y are excluded).

Question 8. Raghu wishes to buy a laptop. He can buy it by paying Rs.40,000 cash or by giving it in 10 installments as Rs.4800 in the first month, Rs.4750 in the second month, Rs.4700 in the third month and so on. If he pays the money in this fashion, find
(i) total amount paid in 10 installments.
(ii) how much extra amount that he has to pay than the cost?
Answer:
The installment payments form an A.P.: 4800, 4750, 4700, ...
The first term \( a = 4800 \).
The common difference \( d = 4750 - 4800 = -50 \).
The number of installments \( n = 10 \).

(i) **Total amount paid in 10 installments.**
We use the sum formula for an A.P.: \( S_n = \frac { n }{ 2 } [2a + (n - 1) d] \).
Substitute the values:
\( S_{10} = \frac { 10 }{ 2 } [2(4800) + (10 - 1)(-50)] \)
\( S_{10} = 5 [9600 + 9(-50)] \)
\( S_{10} = 5 [9600 - 450] \)
\( S_{10} = 5 [9150] \)
\( S_{10} = 45750 \)
The total amount paid in 10 installments is Rs.45,750.

(ii) **How much extra amount that he has to pay than the cash cost.**
Cash cost of the laptop = Rs.40,000.
Total amount paid in installments = Rs.45,750.
Extra amount paid = Total amount paid in installments - Cash cost
Extra amount paid = Rs.45,750 - Rs.40,000
Extra amount paid = Rs.5,750.
In simple words: First, we treat the monthly payments as a list of numbers (an AP) to find the total sum paid over 10 months. The payments decrease each month, so the common difference is negative. Once we have the total paid, we subtract the original cash price of the laptop to find out how much extra money Raghu paid by choosing the installment plan. This extra money represents the cost of paying in installments.

🎯 Exam Tip: When dealing with installment problems, check if payments form an A.P. or G.P. Identify the first term, common difference (or ratio), and number of terms to correctly calculate the total amount paid. Remember to convert rupees symbol to "Rs."

Question 9. A man repays a loan of Rs.65,000 by paying Rs.400 in the first month and then increasing the payment by Rs.300 every month. How long will it take for him to clear the loan?
Answer:
Total loan amount \( S_n = \text{Rs.65,000} \).
First month's payment \( a = \text{Rs.400} \).
Monthly increase in payment \( d = \text{Rs.300} \).
Let \( n \) be the number of months it takes to clear the loan.
The payments form an A.P., so we use the sum formula: \( S_n = \frac { n }{ 2 } [2a + (n - 1) d] \).
Substitute the known values:
\( 65000 = \frac { n }{ 2 } [2(400) + (n - 1)300] \)
Multiply both sides by 2:
\( 130000 = n [800 + 300n - 300] \)
\( 130000 = n [500 + 300n] \)
\( 130000 = 500n + 300n^2 \)
Rearrange into a standard quadratic equation:
\( 300n^2 + 500n - 130000 = 0 \)
Divide the entire equation by 100 to simplify:
\( 3n^2 + 5n - 1300 = 0 \)
Now, we factorize the quadratic equation. We need two numbers that multiply to \( 3 \times (-1300) = -3900 \) and add up to 5. These numbers are 65 and -60.
\( 3n^2 + 65n - 60n - 1300 = 0 \)
Group the terms:
\( n(3n + 65) - 20(3n + 65) = 0 \)
Factor out the common term \( (3n + 65) \):
\( (3n + 65)(n - 20) = 0 \)
This gives two possible values for \( n \):
\( 3n + 65 = 0 \implies 3n = -65 \implies n = -\frac{65}{3} \)
\( n - 20 = 0 \implies n = 20 \)
Since the number of months cannot be negative, we discard \( n = -\frac{65}{3} \).
Therefore, the time taken to pay the loan is 20 months.
In simple words: The man's payments form a pattern where he pays a certain amount, and then pays more each month. This is a special kind of list called an arithmetic progression. We use the formula for the total sum of such a list to find how many months (n) it will take for the total payments to reach the loan amount. We solve the equation for 'n' and pick the positive answer, because time cannot be negative.

🎯 Exam Tip: Loan repayment problems where payments change by a fixed amount each period are typically A.P. questions. Form a quadratic equation for \( n \) and always choose the positive integer solution, as \( n \) (number of periods) must be positive.

Question 10. A brick staircase has a total of 30 steps. The bottom step requires 100 bricks. Each successive step requires two bricks less than the previous step.
(i) How many bricks are required for the top most step?
(ii) How many bricks are required to build the stair case?
Answer:
The number of bricks for each step forms an A.P.
The first term (bricks for the bottom step) \( a = 100 \).
The common difference (bricks decrease by 2 each step) \( d = -2 \).
The total number of steps \( n = 30 \).

(i) **Bricks required for the top most step (\( t_{30} \)).**
We use the formula for the \( n^{th} \) term of an A.P.: \( t_n = a + (n - 1)d \).
Substitute the values:
\( t_{30} = 100 + (30 - 1)(-2) \)
\( t_{30} = 100 + 29(-2) \)
\( t_{30} = 100 - 58 \)
\( t_{30} = 42 \)
So, 42 bricks are required for the top most step.

(ii) **Total number of bricks required to build the stair case (\( S_{30} \)).**
We use the sum formula for an A.P.: \( S_n = \frac { n }{ 2 } [2a + (n - 1) d] \).
Substitute the values:
\( S_{30} = \frac { 30 }{ 2 } [2(100) + (30 - 1)(-2)] \)
\( S_{30} = 15 [200 + 29(-2)] \)
\( S_{30} = 15 [200 - 58] \)
\( S_{30} = 15 [142] \)
\( S_{30} = 2130 \)
So, 2130 bricks are required to build the entire staircase.
In simple words: This problem is about a decreasing pattern of bricks on a staircase. For part (i), we find the number of bricks needed for the very last step by using the formula for any specific term in a list that follows a pattern. For part (ii), we add up all the bricks needed for every single step from bottom to top, using the sum formula for such a pattern.

🎯 Exam Tip: Problems involving stairs, stacks, or layers often use A.P. formulas. Clearly identify \( a \) (first term), \( d \) (common difference, which can be negative if decreasing), and \( n \) (number of steps/layers).

Question 11. If \( S_1, S_2, S_3, \dots, S_m \) are the sums of \( n \) terms of \( m \) A.P.,s whose first terms are \( 1,2,3,\dots,m \) and whose common differences are \( 1,3,5,\dots,(2m – 1) \) respectively, then show that \( (S_1 + S_2 + S_3 + \dots + S_m) = \frac { 1 }{ 2 } mn(mn + 1) \)
Answer:
We are given \( m \) arithmetic progressions. For each A.P., the sum of \( n \) terms is denoted as \( S_k \), where \( k \) ranges from 1 to \( m \).
For the \( k^{th} \) A.P.:
First term is \( a_k = k \).
Common difference is \( d_k = (2k - 1) \).
Number of terms is \( n \).
The sum of \( n \) terms for the \( k^{th} \) A.P. is \( S_k = \frac{n}{2} [2a_k + (n-1)d_k] \).
Substitute \( a_k = k \) and \( d_k = (2k - 1) \):
\( S_k = \frac{n}{2} [2k + (n-1)(2k - 1)] \)
\( S_k = \frac{n}{2} [2k + 2nk - n - 2k + 1] \)
\( S_k = \frac{n}{2} [2nk - n + 1] \)
Now, we need to find the sum of all these \( S_k \) from \( k=1 \) to \( k=m \):
\( \sum_{k=1}^{m} S_k = \sum_{k=1}^{m} \frac{n}{2} [2nk - n + 1] \)
Since \( \frac{n}{2} \) is constant with respect to \( k \), we can take it out:
\( \sum_{k=1}^{m} S_k = \frac{n}{2} \sum_{k=1}^{m} [2nk - n + 1] \)
Split the sum into parts:
\( \sum_{k=1}^{m} S_k = \frac{n}{2} \left[ \sum_{k=1}^{m} (2nk) - \sum_{k=1}^{m} n + \sum_{k=1}^{m} 1 \right] \)
For the first term, \( 2n \) is constant with respect to \( k \):
\( \sum_{k=1}^{m} (2nk) = 2n \sum_{k=1}^{m} k = 2n \left( \frac{m(m+1)}{2} \right) = nm(m+1) \)
For the second term, \( n \) is constant with respect to \( k \):
\( \sum_{k=1}^{m} n = n \times m = nm \)
For the third term, 1 is constant with respect to \( k \):
\( \sum_{k=1}^{m} 1 = 1 \times m = m \)
Substitute these back into the sum expression:
\( \sum_{k=1}^{m} S_k = \frac{n}{2} [nm(m+1) - nm + m] \)
\( \sum_{k=1}^{m} S_k = \frac{n}{2} [nm^2 + nm - nm + m] \)
\( \sum_{k=1}^{m} S_k = \frac{n}{2} [nm^2 + m] \)
Factor out \( m \) from the bracket:
\( \sum_{k=1}^{m} S_k = \frac{n}{2} m [nm + 1] \)
\( \sum_{k=1}^{m} S_k = \frac{1}{2} mn(mn + 1) \)
Thus, \( (S_1 + S_2 + S_3 + \dots + S_m) = \frac { 1 }{ 2 } mn(mn + 1) \).
In simple words: We have 'm' different lists of numbers (AP series), and each list has 'n' terms. The starting number and the common difference for each list change based on its position in the 'm' lists. We first find a general formula for the sum of 'n' terms for any one of these 'm' lists. Then, we add up all these sums together. By simplifying the resulting expression, we show that it equals the target formula. This involves using sum properties like \( \sum k = \frac{m(m+1)}{2} \).

🎯 Exam Tip: When dealing with sums of sums, first find a general expression for \( S_k \) (the sum of \( n \) terms for the \( k^{th} \) AP). Then, sum these \( S_k \) expressions from \( k=1 \) to \( m \), using known summation formulas like \( \sum k \) and \( \sum (\text{constant}) \).

Question 12. Find the sum \( \left[ \frac{a-b}{a+b} + \frac{3a-2b}{a+b} + \frac{5a-3b}{a+b} + \dots \text{to 12 terms} \right] \)
Answer:
The given series is \( \frac{a-b}{a+b} + \frac{3a-2b}{a+b} + \frac{5a-3b}{a+b} + \dots \)
Let's identify if this is an A.P. The terms have a common denominator \( (a+b) \).
The first term is \( t_1 = \frac{a-b}{a+b} \).
The second term is \( t_2 = \frac{3a-2b}{a+b} \).
The common difference \( d = t_2 - t_1 \):
\( d = \frac{3a-2b}{a+b} - \frac{a-b}{a+b} \)
\( d = \frac{(3a-2b) - (a-b)}{a+b} \)
\( d = \frac{3a-2b - a + b}{a+b} \)
\( d = \frac{2a-b}{a+b} \)
We need to find the sum of the series to 12 terms, so \( n = 12 \).
The formula for the sum of \( n \) terms of an A.P. is \( S_n = \frac{n}{2} [2t_1 + (n-1)d] \).
Substitute the values for \( n \), \( t_1 \), and \( d \):
\( S_{12} = \frac{12}{2} \left[ 2 \left( \frac{a-b}{a+b} \right) + (12-1) \left( \frac{2a-b}{a+b} \right) \right] \)
\( S_{12} = 6 \left[ \frac{2(a-b)}{a+b} + 11 \left( \frac{2a-b}{a+b} \right) \right] \)
\( S_{12} = 6 \left[ \frac{2a-2b}{a+b} + \frac{22a-11b}{a+b} \right] \)
\( S_{12} = 6 \left[ \frac{(2a-2b) + (22a-11b)}{a+b} \right] \)
\( S_{12} = 6 \left[ \frac{2a+22a - 2b-11b}{a+b} \right] \)
\( S_{12} = 6 \left[ \frac{24a-13b}{a+b} \right] \)
So, the sum of the series to 12 terms is \( S_{12} = \frac{6(24a-13b)}{a+b} \).
In simple words: We are asked to find the sum of a list of terms. First, we check if it's an Arithmetic Progression (AP) by finding the difference between the first two terms. This difference, called the common difference, is found to be \( \frac{2a-b}{a+b} \). Then, we use the sum formula for an AP, putting in the first term, the common difference, and the number of terms (which is 12). We simplify the expression to get the final sum.

🎯 Exam Tip: When terms are fractions with a common denominator, you can treat the numerators as an A.P. to find the common difference. Always simplify algebraic expressions carefully when adding or subtracting fractions.