RBSE Solutions Class 11 Maths Chapter 8 Sequence, Progression, and Series Exercise 8.2

 

Question 1. Find the sum of the following progression :
(i) \( 7+11+15+19+ \dots \) upto 20 terms
(ii) \( \frac{1}{3} + 1 + \frac{5}{3} + \dots \) upto 10 terms
(iii) \( \frac{1}{\sqrt{2}+1} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}-1} + \dots \) upto 6 terms
Answer:
(i) Given progression: \( 7+11+15+19+ \dots \).
Here, the first term \( a = 7 \), the common difference \( d = 11 - 7 = 4 \), and the number of terms \( n = 20 \).
The formula for the sum of an arithmetic progression is \( S_n = \frac{n}{2}[2a + (n-1)d] \).
Now, we put the values into the formula:
\( S_{20} = \frac{20}{2}[2 \times 7 + (20-1) \times 4] \)
\( S_{20} = 10[14 + 19 \times 4] \)
\( S_{20} = 10[14 + 76] \)
\( S_{20} = 10[90] \)
\( S_{20} = 900 \)
So, the sum of the first 20 terms is 900.
(ii) Given progression: \( \frac{1}{3} + 1 + \frac{5}{3} + \dots \).
Here, the first term \( a = \frac{1}{3} \).
The common difference \( d = 1 - \frac{1}{3} = \frac{3-1}{3} = \frac{2}{3} \).
The number of terms \( n = 10 \).
Using the formula for the sum of an arithmetic progression:
\( S_n = \frac{n}{2}[2a + (n-1)d] \)
Now, we substitute the values:
\( S_{10} = \frac{10}{2}[2 \times \frac{1}{3} + (10-1) \times \frac{2}{3}] \)
\( S_{10} = 5[\frac{2}{3} + 9 \times \frac{2}{3}] \)
\( S_{10} = 5[\frac{2}{3} + \frac{18}{3}] \)
\( S_{10} = 5[\frac{2+18}{3}] \)
\( S_{10} = 5[\frac{20}{3}] \)
\( S_{10} = \frac{100}{3} \)
Therefore, the sum of the first 10 terms is \( \frac{100}{3} \).
(iii) Given progression: \( \frac{1}{\sqrt{2}+1} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}-1} + \dots \).
First, simplify the first term by rationalizing the denominator:
\( a = \frac{1}{\sqrt{2}+1} = \frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{\sqrt{2}-1}{(\sqrt{2})^2 - 1^2} = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1 \)
So, the first term \( a = \sqrt{2}-1 \).
Next, calculate the common difference \( d \). The second term is \( \frac{1}{\sqrt{2}} \) and the third term is \( \frac{1}{\sqrt{2}-1} \). Let's use the given first term \( a \) and the second term \( \frac{1}{\sqrt{2}} \) to find \( d \).
\( d = \text{second term} - \text{first term} \)
\( d = \frac{1}{\sqrt{2}} - (\sqrt{2}-1) = \frac{1}{\sqrt{2}} - \sqrt{2} + 1 = \frac{1 - (\sqrt{2})^2}{\sqrt{2}} + 1 = \frac{1 - 2}{\sqrt{2}} + 1 = \frac{-1}{\sqrt{2}} + 1 = 1 - \frac{1}{\sqrt{2}} \)
Let's check with the third term as well, to confirm it's an A.P. If the common difference is consistent, then \( \frac{1}{\sqrt{2}-1} - \frac{1}{\sqrt{2}} \) should be equal to \( d \).
\( \frac{1}{\sqrt{2}-1} = \frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)} = \frac{\sqrt{2}+1}{2-1} = \sqrt{2}+1 \)
So the progression is \( (\sqrt{2}-1) + \frac{1}{\sqrt{2}} + (\sqrt{2}+1) + \dots \).
\( d = \frac{1}{\sqrt{2}} - (\sqrt{2}-1) = 1 - \frac{1}{\sqrt{2}} \).
The number of terms \( n = 6 \).
The formula for the sum of an arithmetic progression is \( S_n = \frac{n}{2}[2a + (n-1)d] \).
Substitute the values:
\( S_6 = \frac{6}{2}[2(\sqrt{2}-1) + (6-1)(1-\frac{1}{\sqrt{2}})] \)
\( S_6 = 3[2\sqrt{2}-2 + 5(1-\frac{1}{\sqrt{2}})] \)
\( S_6 = 3[2\sqrt{2}-2 + 5 - \frac{5}{\sqrt{2}}] \)
\( S_6 = 3[3 + 2\sqrt{2} - \frac{5}{\sqrt{2}}] \)
\( S_6 = 3[3 + \frac{2\sqrt{2} \times \sqrt{2} - 5}{\sqrt{2}}] \)
\( S_6 = 3[3 + \frac{4-5}{\sqrt{2}}] \)
\( S_6 = 3[3 - \frac{1}{\sqrt{2}}] \)
\( S_6 = 3[\frac{3\sqrt{2}-1}{\sqrt{2}}] \)
\( S_6 = \frac{3(3\sqrt{2}-1)}{\sqrt{2}} \)
To rationalize the denominator, multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \):
\( S_6 = \frac{3(3\sqrt{2}-1)\sqrt{2}}{\sqrt{2}\sqrt{2}} = \frac{3(3 \times 2 - \sqrt{2})}{2} = \frac{3(6-\sqrt{2})}{2} \)
Therefore, the sum of 6 terms is \( \frac{3(6-\sqrt{2})}{2} \).
In simple words: For each series, we first find the starting number (first term) and the constant difference between numbers. Then, we use a special formula that quickly adds up all the numbers in the series, up to a certain number of terms. This saves us from adding them one by one.

🎯 Exam Tip: Always identify the type of progression (Arithmetic Progression in this case) and correctly find the first term (a), common difference (d), and number of terms (n) before applying the sum formula.

 

Question 2. Find the sum of all odd integers from 1 to 101, which is divisible by 3.
Answer: We need to find odd integers between 1 and 101 that are divisible by 3.
The first odd integer greater than 1 that is divisible by 3 is 3 itself.
The next such number is 9 (not 6, because 6 is even). Then 15, and so on.
This forms an arithmetic progression (A.P.): 3, 9, 15, ..., 99.
Here, the first term \( a = 3 \).
The common difference \( d = 9 - 3 = 6 \).
The last term \( l = 99 \).
We need to find the number of terms \( n \) using the formula for the nth term: \( l = a + (n-1)d \).
\( 99 = 3 + (n-1)6 \)
\( 99 - 3 = (n-1)6 \)
\( 96 = (n-1)6 \)
\( \frac{96}{6} = n-1 \)
\( 16 = n-1 \)
\( n = 16 + 1 \)
\( n = 17 \)
Now, we find the sum of these 17 terms using the formula \( S_n = \frac{n}{2}[a+l] \), or \( S_n = \frac{n}{2}[2a + (n-1)d] \). Let's use the first one as we have \( a \) and \( l \).
\( S_{17} = \frac{17}{2}[3 + 99] \)
\( S_{17} = \frac{17}{2}[102] \)
\( S_{17} = 17 \times \frac{102}{2} \)
\( S_{17} = 17 \times 51 \)
\( S_{17} = 867 \)
Thus, the sum of all odd integers from 1 to 101 which are divisible by 3 is 867.
In simple words: We are looking for odd numbers between 1 and 101 that can be divided by 3. These numbers form a pattern. We find the first and last numbers in this pattern, count how many numbers there are, and then use a quick sum formula to add them all up.

🎯 Exam Tip: When finding numbers divisible by a certain value within a range, make sure to consider all conditions (like 'odd' or 'even') carefully when determining the first term and common difference.

 

Question 3. Find the sum of n terms of A.P. whose rth term is 2r + 3.
Answer: We are given that the rth term of an A.P. is \( T_r = 2r + 3 \).
To find the sum of \( n \) terms, we first need to find the first term \( a \) and the common difference \( d \).
We can find the first term \( T_1 \) by setting \( r=1 \):
\( T_1 = 2(1) + 3 = 2 + 3 = 5 \). So, the first term \( a = 5 \).
We can find the second term \( T_2 \) by setting \( r=2 \):
\( T_2 = 2(2) + 3 = 4 + 3 = 7 \).
The common difference \( d \) is the difference between consecutive terms:
\( d = T_2 - T_1 = 7 - 5 = 2 \).
Now we have the first term \( a = 5 \), the common difference \( d = 2 \), and the number of terms is \( n \).
The formula for the sum of \( n \) terms of an A.P. is \( S_n = \frac{n}{2}[2a + (n-1)d] \).
Substitute the values of \( a \) and \( d \) into the formula:
\( S_n = \frac{n}{2}[2(5) + (n-1)2] \)
\( S_n = \frac{n}{2}[10 + 2n - 2] \)
\( S_n = \frac{n}{2}[8 + 2n] \)
We can factor out 2 from the bracket:
\( S_n = \frac{n}{2} \times 2[4 + n] \)
\( S_n = n(n+4) \)
So, the sum of \( n \) terms of this A.P. is \( n(n+4) \). This makes sense because the sum of terms in an AP is always a quadratic expression in n.
In simple words: We know a rule for finding any term in the number pattern. Using this rule, we find the first two numbers to see how the pattern starts and how much it increases each time. Once we know this, we use the sum formula to find the total of any number of terms.

🎯 Exam Tip: Remember that if the nth term formula is given, you can always find the first term (a) by substituting n=1, and the common difference (d) by finding T2-T1. These are essential for finding the sum.

 

Question 5. If sum of n terms of A.P. 1, 6, 11,.... is 148, then find number of terms and last term.
Answer: Given an arithmetic progression: 1, 6, 11, ....
Here, the first term \( a = 1 \).
The common difference \( d = 6 - 1 = 5 \).
The sum of \( n \) terms \( S_n = 148 \).
We use the formula for the sum of \( n \) terms of an A.P.: \( S_n = \frac{n}{2}[2a + (n-1)d] \).
Substitute the given values into the formula:
\( 148 = \frac{n}{2}[2(1) + (n-1)5] \)
\( 148 \times 2 = n[2 + 5n - 5] \)
\( 296 = n[5n - 3] \)
\( 296 = 5n^2 - 3n \)
Rearrange this into a quadratic equation:
\( 5n^2 - 3n - 296 = 0 \)
We can solve this quadratic equation using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a=5, b=-3, c=-296 \).
\( n = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(5)(-296)}}{2(5)} \)
\( n = \frac{3 \pm \sqrt{9 + 5920}}{10} \)
\( n = \frac{3 \pm \sqrt{5929}}{10} \)
\( n = \frac{3 \pm 77}{10} \)
Now, we calculate the two possible values for \( n \):
Taking the positive sign: \( n = \frac{3 + 77}{10} = \frac{80}{10} = 8 \).
Taking the negative sign: \( n = \frac{3 - 77}{10} = \frac{-74}{10} = -7.4 \).
Since the number of terms \( n \) cannot be negative or a fraction, we discard \( n = -7.4 \).
So, the number of terms is \( n = 8 \).
Now we need to find the last term, \( T_n \), which is \( T_8 \).
Use the formula for the nth term of an A.P.: \( T_n = a + (n-1)d \).
\( T_8 = 1 + (8-1)5 \)
\( T_8 = 1 + (7)5 \)
\( T_8 = 1 + 35 \)
\( T_8 = 36 \)
Therefore, the number of terms is 8 and the last term is 36.
In simple words: We have a number pattern and its total sum. We need to find how many numbers are in the pattern and what the last number is. We use a formula for the sum that gives us an equation. Solving this equation tells us the count of numbers, and then we find the last number using another formula.

🎯 Exam Tip: When solving for the number of terms (n), remember that n must always be a positive whole number. Discard any fractional or negative solutions. Also, double-check your calculations in the quadratic formula.

 

Question 7. If sum of n, 2n, 3n terms of any A.P. are S1, S2 and S3 respectively, then prove that S3 = 3 (S2 – S1).
Answer: Let \( a \) be the first term and \( d \) be the common difference of the arithmetic progression.
The sum of \( n \) terms is \( S_n = \frac{n}{2}[2a + (n-1)d] \).
Given:
Sum of \( n \) terms: \( S_1 = \frac{n}{2}[2a + (n-1)d] \) ...(i)
Sum of \( 2n \) terms: \( S_2 = \frac{2n}{2}[2a + (2n-1)d] \) ...(ii)
Sum of \( 3n \) terms: \( S_3 = \frac{3n}{2}[2a + (3n-1)d] \) ...(iii)
Now, let's find \( S_2 - S_1 \):
\( S_2 - S_1 = \frac{2n}{2}[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d] \)
Factor out \( \frac{n}{2} \):
\( S_2 - S_1 = \frac{n}{2} \left( 2[2a + (2n-1)d] - [2a + (n-1)d] \right) \)
\( S_2 - S_1 = \frac{n}{2} \left( [4a + (4n-2)d] - [2a + (n-1)d] \right) \)
\( S_2 - S_1 = \frac{n}{2} \left( 4a + 4nd - 2d - 2a - nd + d \right) \)
Combine like terms inside the bracket:
\( S_2 - S_1 = \frac{n}{2} \left( (4a-2a) + (4nd-nd) + (-2d+d) \right) \)
\( S_2 - S_1 = \frac{n}{2} [2a + 3nd - d] \)
\( S_2 - S_1 = \frac{n}{2} [2a + (3n-1)d] \) ...(iv)
Now, we need to prove that \( S_3 = 3(S_2 - S_1) \).
Multiply the expression for \( S_2 - S_1 \) (equation iv) by 3:
\( 3(S_2 - S_1) = 3 \times \frac{n}{2} [2a + (3n-1)d] \)
\( 3(S_2 - S_1) = \frac{3n}{2} [2a + (3n-1)d] \)
By comparing this with equation (iii), we see that:
\( 3(S_2 - S_1) = S_3 \)
Hence, it is proven that \( S_3 = 3(S_2 - S_1) \). This property shows a consistent relationship between sums of terms in an arithmetic progression.
In simple words: We write down the formulas for the sum of 'n' terms, '2n' terms, and '3n' terms in a number pattern. Then we work out the difference between the sum of '2n' terms and 'n' terms. Finally, we show that if we multiply this difference by 3, we get exactly the sum of '3n' terms.

🎯 Exam Tip: Remember the general formula for the sum of an A.P. and practice algebraic manipulation to simplify expressions. Clearly label each sum (S1, S2, S3) and derive expressions for each before performing the required operations.

 

Question 10. Find three numbers in A.P., whose sum is 12 and sum of their cube is 408.
Answer: Let the three numbers in arithmetic progression be \( a-d \), \( a \), and \( a+d \).
Given that their sum is 12:
\( (a-d) + a + (a+d) = 12 \)
\( 3a = 12 \)
\( a = \frac{12}{3} \)
\( a = 4 \)
So, the middle term is 4.
Given that the sum of their cubes is 408:
\( (a-d)^3 + a^3 + (a+d)^3 = 408 \)
Substitute \( a=4 \):
\( (4-d)^3 + 4^3 + (4+d)^3 = 408 \)
Expand the cubic terms using \( (x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3 \) and \( (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 \):
\( (4^3 - 3 \cdot 4^2 \cdot d + 3 \cdot 4 \cdot d^2 - d^3) + 4^3 + (4^3 + 3 \cdot 4^2 \cdot d + 3 \cdot 4 \cdot d^2 + d^3) = 408 \)
\( (64 - 48d + 12d^2 - d^3) + 64 + (64 + 48d + 12d^2 + d^3) = 408 \)
Notice that the terms \( -48d \) and \( +48d \) cancel out, and \( -d^3 \) and \( +d^3 \) also cancel out.
\( 64 + 12d^2 + 64 + 64 + 12d^2 = 408 \)
Combine the constant terms and the \( d^2 \) terms:
\( (64+64+64) + (12d^2+12d^2) = 408 \)
\( 192 + 24d^2 = 408 \)
Subtract 192 from both sides:
\( 24d^2 = 408 - 192 \)
\( 24d^2 = 216 \)
Divide by 24:
\( d^2 = \frac{216}{24} \)
\( d^2 = 9 \)
Take the square root of both sides:
\( d = \pm \sqrt{9} \)
\( d = \pm 3 \)
If \( d=3 \), the numbers are \( a-d = 4-3=1 \), \( a=4 \), \( a+d=4+3=7 \). The sequence is 1, 4, 7.
If \( d=-3 \), the numbers are \( a-d = 4-(-3)=7 \), \( a=4 \), \( a+d=4+(-3)=1 \). The sequence is 7, 4, 1.
Both sequences represent the same three numbers in A.P., just in a different order. Therefore, the three numbers are 1, 4, and 7.
In simple words: We imagine the three numbers in a pattern where the middle number is 'a' and the other two are 'a-d' and 'a+d'. First, we use their total sum to find the middle number. Then, we use the sum of their cubes to find the common difference. This gives us the three numbers.

🎯 Exam Tip: When dealing with an odd number of terms in an A.P., it's usually easiest to choose the terms as \( a-d, a, a+d \) (for 3 terms) or \( a-2d, a-d, a, a+d, a+2d \) (for 5 terms). This simplifies the sum, allowing \( d \) to cancel out and directly find \( a \).

 

Question 11. If n arithmetic mean are inserted in between 1 and 51 such that ratio of 4th and 7th arithmetic mean is 3:5 then find the value of n.
Answer: Let the two numbers be \( A=1 \) and \( B=51 \). We insert \( n \) arithmetic means between them. So the sequence is \( A, A_1, A_2, \dots, A_n, B \).
The total number of terms in this arithmetic progression is \( n+2 \).
The first term is \( a = 1 \).
The last term is \( l = B = 51 \). This is the \( (n+2) \)-th term.
Using the formula for the last term, \( l = a + (\text{total terms}-1)d \):
\( 51 = 1 + (n+2-1)d \)
\( 51 = 1 + (n+1)d \)
\( 50 = (n+1)d \)
\( d = \frac{50}{n+1} \)
The k-th arithmetic mean \( A_k \) inserted between \( a \) and \( b \) is given by \( A_k = a + kd \).
The 4th arithmetic mean is \( A_4 = a + 4d \).
Substitute \( a=1 \) and \( d=\frac{50}{n+1} \):
\( A_4 = 1 + 4 \left( \frac{50}{n+1} \right) = 1 + \frac{200}{n+1} = \frac{n+1+200}{n+1} = \frac{n+201}{n+1} \)
The 7th arithmetic mean is \( A_7 = a + 7d \).
Substitute \( a=1 \) and \( d=\frac{50}{n+1} \):
\( A_7 = 1 + 7 \left( \frac{50}{n+1} \right) = 1 + \frac{350}{n+1} = \frac{n+1+350}{n+1} = \frac{n+351}{n+1} \)
We are given that the ratio of the 4th and 7th arithmetic mean is 3:5:
\( \frac{A_4}{A_7} = \frac{3}{5} \)
Substitute the expressions for \( A_4 \) and \( A_7 \):
\( \frac{\frac{n+201}{n+1}}{\frac{n+351}{n+1}} = \frac{3}{5} \)
The \( (n+1) \) terms cancel out:
\( \frac{n+201}{n+351} = \frac{3}{5} \)
Now, cross-multiply to solve for \( n \):
\( 5(n+201) = 3(n+351) \)
\( 5n + 1005 = 3n + 1053 \)
Subtract \( 3n \) from both sides:
\( 2n + 1005 = 1053 \)
Subtract 1005 from both sides:
\( 2n = 1053 - 1005 \)
\( 2n = 48 \)
Divide by 2:
\( n = \frac{48}{2} \)
\( n = 24 \)
Therefore, the value of \( n \) is 24. This means 24 arithmetic means are inserted between 1 and 51.
In simple words: We are placing 'n' numbers evenly between 1 and 51. We found a way to write any of these inserted numbers using 'n'. Then, we used the given ratio of the 4th and 7th inserted numbers to set up an equation. Solving this equation helped us find the value of 'n'.

🎯 Exam Tip: Remember that if 'n' arithmetic means are inserted between 'a' and 'b', the total number of terms in the A.P. is 'n+2'. Also, the k-th arithmetic mean is found by \( A_k = a+kd \), where \( a \) is the first term of the sequence.

 

Question 12. If x, y, z are in A.P., then prove that :
(i) y + z, z + x, x + y are in A.P.
(ii) \( \frac{1}{yz}, \frac{1}{zx}, \frac{1}{xy} \) are in A.P.
(iii) \( (x - y)(y - z) = \frac{(z-x)^2}{4} \)
(iv) \( (x - z)^2 = 4(y^2 - xz) \)
(v) \( xy + yz + zx = \frac{x^2 + z^2 + 4xz}{2} \)
Answer: If \( x, y, z \) are in A.P., then by definition, \( 2y = x + z \). We will use this condition to prove each part.
(i) To prove \( y+z, z+x, x+y \) are in A.P., we must show that the middle term is the average of the other two, i.e., \( 2(z+x) = (y+z) + (x+y) \).
Right Hand Side (RHS): \( (y+z) + (x+y) = x + 2y + z \).
Since \( 2y = x+z \), substitute this into RHS:
RHS: \( x + (x+z) + z = 2x + 2z = 2(x+z) \).
Left Hand Side (LHS): \( 2(z+x) = 2(x+z) \).
Since LHS = RHS, \( y+z, z+x, x+y \) are in A.P.
(ii) To prove \( \frac{1}{yz}, \frac{1}{zx}, \frac{1}{xy} \) are in A.P., we must show that \( 2 \left( \frac{1}{zx} \right) = \frac{1}{yz} + \frac{1}{xy} \).
RHS: \( \frac{1}{yz} + \frac{1}{xy} = \frac{x}{xyz} + \frac{z}{xyz} = \frac{x+z}{xyz} \).
Since \( x, y, z \) are in A.P., we know \( x+z = 2y \). Substitute this into RHS:
RHS: \( \frac{2y}{xyz} = \frac{2}{xz} \).
LHS: \( 2 \left( \frac{1}{zx} \right) = \frac{2}{zx} \).
Since LHS = RHS, \( \frac{1}{yz}, \frac{1}{zx}, \frac{1}{xy} \) are in A.P.
(iii) To prove \( (x - y)(y - z) = \frac{(z-x)^2}{4} \).
Since \( x, y, z \) are in A.P., let \( y-x = d \) and \( z-y = d \), where \( d \) is the common difference.
This means \( y = x+d \) and \( z = y+d = x+2d \).
LHS: \( (x-y)(y-z) \).
Substitute \( y=x+d \) and \( z=x+2d \):
\( (x - (x+d))((x+d) - (x+2d)) \)
\( (-d)(-d) = d^2 \).
RHS: \( \frac{(z-x)^2}{4} \).
Substitute \( z=x+2d \):
\( \frac{((x+2d)-x)^2}{4} = \frac{(2d)^2}{4} = \frac{4d^2}{4} = d^2 \).
Since LHS = RHS, the statement is proven.
(iv) To prove \( (x - z)^2 = 4(y^2 - xz) \).
Since \( x, y, z \) are in A.P., \( 2y = x+z \), so \( y = \frac{x+z}{2} \).
LHS: \( (x-z)^2 \).
RHS: \( 4(y^2 - xz) \).
Substitute \( y = \frac{x+z}{2} \) into RHS:
\( 4 \left( \left(\frac{x+z}{2}\right)^2 - xz \right) \)
\( 4 \left( \frac{x^2+2xz+z^2}{4} - xz \right) \)
\( 4 \left( \frac{x^2+2xz+z^2 - 4xz}{4} \right) \)
\( 4 \left( \frac{x^2-2xz+z^2}{4} \right) \)
\( x^2-2xz+z^2 \)
This is equal to \( (x-z)^2 \).
Since LHS = RHS, the statement is proven.
(v) To prove \( xy + yz + zx = \frac{x^2 + z^2 + 4xz}{2} \).
Since \( x, y, z \) are in A.P., \( y = \frac{x+z}{2} \).
LHS: \( xy + yz + zx \).
Factor out \( y \) from the first two terms:
\( y(x+z) + zx \).
Substitute \( y = \frac{x+z}{2} \):
\( \frac{x+z}{2}(x+z) + zx \)
\( \frac{(x+z)^2}{2} + zx \)
\( \frac{x^2+2xz+z^2}{2} + zx \)
To add the terms, make the denominator common:
\( \frac{x^2+2xz+z^2}{2} + \frac{2xz}{2} \)
\( \frac{x^2+2xz+z^2+2xz}{2} \)
\( \frac{x^2+z^2+4xz}{2} \)
This is equal to the RHS. Since LHS = RHS, the statement is proven.
In simple words: When three numbers are in an arithmetic progression, the middle number is always the average of the first and last numbers. We use this key rule to show that various combinations of these numbers also follow the same pattern or satisfy certain equations. It confirms the consistent nature of arithmetic progressions.

🎯 Exam Tip: For problems involving proofs with arithmetic progressions, always start by stating the fundamental property: if x, y, z are in A.P., then \( 2y = x+z \). This relationship is the key to simplifying expressions and proving identities.

 

Question 14. Find the sum of A.P. a1 + a2 + a3 ..., A30.
Answer: We need to find the sum of 30 terms of an A.P., which is \( S_{30} \).
We are given the condition: \( a_1 + a_7 + a_{10} + a_{21} + a_{24} + a_{30} = 540 \).
In an arithmetic progression, the sum of terms equidistant from the beginning and end is constant and equal to the sum of the first and last terms. That is, \( a_k + a_{n-k+1} = a_1 + a_n \).
For a 30-term A.P. (where \( n=30 \)):
1. \( a_1 + a_{30} \)
2. \( a_7 \). The term corresponding to \( a_7 \) from the end is \( a_{30-7+1} = a_{24} \). So, \( a_7 + a_{24} = a_1 + a_{30} \).
3. \( a_{10} \). The term corresponding to \( a_{10} \) from the end is \( a_{30-10+1} = a_{21} \). So, \( a_{10} + a_{21} = a_1 + a_{30} \).
Now, let's rewrite the given equation using these properties:
\( (a_1 + a_{30}) + (a_7 + a_{24}) + (a_{10} + a_{21}) = 540 \)
Substitute \( (a_1 + a_{30}) \) for each pair:
\( (a_1 + a_{30}) + (a_1 + a_{30}) + (a_1 + a_{30}) = 540 \)
\( 3(a_1 + a_{30}) = 540 \)
Divide by 3 to find the sum of the first and last terms:
\( a_1 + a_{30} = \frac{540}{3} \)
\( a_1 + a_{30} = 180 \)
The formula for the sum of \( n \) terms of an A.P. is \( S_n = \frac{n}{2}(a_1 + a_n) \).
For \( n=30 \):
\( S_{30} = \frac{30}{2}(a_1 + a_{30}) \)
Substitute the value of \( a_1 + a_{30} \):
\( S_{30} = 15(180) \)
\( S_{30} = 2700 \)
Therefore, the sum of the A.P. \( a_1 + a_2 + a_3 + \dots + a_{30} \) is 2700.
In simple words: We are given a sum of certain terms in a number pattern and asked to find the total sum of all 30 terms. We use a special property that pairs of terms equally distant from the start and end add up to the same value. By grouping the given terms this way, we find the sum of the first and last terms. Then, a simple formula gives us the total sum.

🎯 Exam Tip: Always utilize the property that in an A.P., \( a_k + a_{n-k+1} = a_1 + a_n \). This can significantly simplify complex sum problems and is a crucial concept to master for efficient problem-solving.

(The requested content is not present on pages 15-17 of the document. These pages contain only navigational elements, search bars, comment sections, recent posts, and copyright information, all of which are to be ignored according to the content processing rules.)