RBSE Solutions Class 11 Maths Chapter 8 Sequence, Progression, and Series Exercise 8.7

Question 1. Find the term mentioned in following Harmonic Progression –
(i) \( \frac { 1 }{ 2 }, \frac { 1 }{ 5 }, \frac { 1 }{ 8 }, \frac { 1 }{ 11 }, \dots \) 6th term
(ii) \( \frac { 1 }{ 9 }, \frac { 1 }{ 19 }, \frac { 1 }{ 29 }, \frac { 1 }{ 39 }, \dots \) 18th term
(iii) \( \frac { 1 }{ 14 }, \frac { 2 }{ 29 }, \frac { 1 }{ 15 }, \frac { 2 }{ 31 }, \dots \) 10th term
Answer:
(i) For the given Harmonic Progression (H.P.), the corresponding Arithmetic Progression (A.P.) is found by taking the reciprocals of the terms. So, the A.P. is \( 2, 5, 8, 11, \dots \).
Here, the first term \( a = 2 \).
The common difference \( d = 5 - 2 = 3 \).
To find the 6th term (\( T_6 \)) of the A.P., we use the formula \( T_n = a + (n-1)d \).
\( T_6 = a + 5d \)
\( = 2 + 5 \times 3 \)
\( = 2 + 15 \)
\( = 17 \)
The 6th term of the corresponding H.P. is the reciprocal of the 6th term of the A.P.
Hence, the 6th term of the H.P. is \( \frac{1}{17} \).
(ii) For the given Harmonic Progression, the corresponding Arithmetic Progression (A.P.) is found by taking the reciprocals of the terms. So, the A.P. is \( 9, 19, 29, 39, \dots \).
Here, the first term \( a = 9 \).
The common difference \( d = 19 - 9 = 10 \).
To find the 18th term (\( T_{18} \)) of the A.P., we use the formula \( T_n = a + (n-1)d \).
\( T_{18} = a + 17d \)
\( = 9 + 17 \times 10 \)
\( = 9 + 170 \)
\( = 179 \)
The 18th term of the corresponding H.P. is the reciprocal of the 18th term of the A.P.
Hence, the 18th term of the H.P. is \( \frac{1}{179} \).
(iii) For the given Harmonic Progression, we consider the reciprocal sequence. The source transforms this into an A.P. starting with \( 28, 29, 30, 31, \dots \). This A.P. implies a slightly different initial H.P. like \( \frac{1}{28}, \frac{1}{29}, \frac{1}{30}, \frac{1}{31} \) or \( \frac{2}{28}, \frac{2}{29}, \frac{2}{30}, \frac{2}{31} \).
Here, the first term \( a = 28 \).
The common difference \( d = 29 - 28 = 1 \).
To find the 10th term (\( T_{10} \)) of the A.P., we use the formula \( T_n = a + (n-1)d \).
\( T_{10} = a + 9d \)
\( = 28 + 9 \times 1 \)
\( = 28 + 9 \)
\( = 37 \)
Following the logic of the source, the 10th term of the corresponding H.P. is then taken as \( \frac{2}{37} \). This suggests the original H.P. was expressed in terms of \( \frac{2}{\text{denominator}} \) in the A.P. reciprocal form.
In simple words: To find a term in a Harmonic Progression, first change it into an Arithmetic Progression by taking the reciprocal of each term. Then, find the required term in this new A.P. using the formula \( T_n = a + (n-1)d \). Finally, take the reciprocal of that A.P. term to get the answer for the H.P. Remember to carefully check how the reciprocals are formed if the H.P. terms have different numerators.

🎯 Exam Tip: Always convert the H.P. to its corresponding A.P. before calculating any term. Be careful with fractional terms and ensure you take the reciprocal correctly at each step.

Question 2. Find nth term of the following H.P.:
(i) \( \frac { 1 }{ 5 }, \frac { 2 }{ 9 }, \frac { 1 }{ 4 }, \frac { 2 }{ 7 }, \dots \)
(ii) \( \frac { 2 }{ a+b }, \frac { 1 }{ a }, \frac { 2 }{ 3a-b }, \dots \)
Answer:
(i) First, rewrite the given Harmonic Progression (H.P.) terms to have a common numerator to easily find the corresponding Arithmetic Progression (A.P.).
The H.P. is \( \frac{1}{5}, \frac{2}{9}, \frac{1}{4}, \frac{2}{7}, \dots \)
We can rewrite it as \( \frac{2}{10}, \frac{2}{9}, \frac{2}{8}, \frac{2}{7}, \dots \).
The corresponding A.P. is formed by taking the reciprocals of the denominators (since the numerator is common). So, the A.P. is \( 10, 9, 8, 7, \dots \).
Here, the first term \( a = 10 \).
The common difference \( d = 9 - 10 = -1 \).
To find the \( n^{th} \) term (\( T_n \)) of the A.P., we use the formula \( T_n = a + (n-1)d \).
\( T_n = 10 + (n-1)(-1) \)
\( = 10 - n + 1 \)
\( = 11 - n \)
The \( n^{th} \) term of the corresponding H.P. is the reciprocal of the \( n^{th} \) term of the A.P., with the original common numerator.
Hence, the \( n^{th} \) term of the H.P. is \( \frac{2}{11-n} \).
(ii) For the given Harmonic Progression (H.P.), we can also rewrite its terms to have a common numerator to simplify finding the corresponding Arithmetic Progression (A.P.).
The H.P. is \( \frac{2}{a+b}, \frac{1}{a}, \frac{2}{3a-b}, \dots \)
We can rewrite it as \( \frac{2}{a+b}, \frac{2}{2a}, \frac{2}{3a-b}, \dots \).
The corresponding A.P. is formed by taking the reciprocals of the denominators. So, the A.P. is \( a+b, 2a, 3a-b, \dots \).
Here, the first term of the A.P. is \( a_1 = a+b \).
The common difference \( d = 2a - (a+b) = 2a - a - b = a-b \).
To find the \( n^{th} \) term (\( T_n \)) of this A.P., we use the formula \( T_n = a_1 + (n-1)d \).
\( T_n = (a+b) + (n-1)(a-b) \)
\( T_n = a+b + na - nb - a + b \)
\( T_n = na - nb + 2b \)
\( T_n = n(a-b) + 2b \)
The \( n^{th} \) term of the corresponding H.P. is the reciprocal of the \( n^{th} \) term of the A.P., using the common numerator of 2.
Hence, the \( n^{th} \) term of the H.P. is \( \frac{2}{n(a-b) + 2b} \).
In simple words: When finding the \( n^{th} \) term of an H.P., first make all numerators the same if they aren't. Then, make an A.P. from the denominators. Calculate the \( n^{th} \) term of this A.P. Finally, write the original common numerator over the \( n^{th} \) term of the A.P. to get the H.P. term.

🎯 Exam Tip: When the numerators are not identical in the H.P., multiply or divide terms to make them uniform. This makes it easier to form the corresponding A.P. from the denominators.

Question 3. Find that H.P. whose 2nd term is \( \frac{2}{5} \) and 7th term is \( \frac{4}{25} \)
Answer:
If the 2nd term of the Harmonic Progression (H.P.) is \( \frac{2}{5} \), then the 2nd term of its corresponding Arithmetic Progression (A.P.) is the reciprocal, \( T_2 = \frac{5}{2} \).
So, \( a + d = \frac{5}{2} \) (Equation 1)
If the 7th term of the H.P. is \( \frac{4}{25} \), then the 7th term of its corresponding A.P. is the reciprocal, \( T_7 = \frac{25}{4} \).
So, \( a + 6d = \frac{25}{4} \) (Equation 2)
Now, we solve these two linear equations to find \( a \) (the first term of the A.P.) and \( d \) (the common difference of the A.P.).
Subtract Equation 1 from Equation 2:
\( (a + 6d) - (a + d) = \frac{25}{4} - \frac{5}{2} \)
\( 5d = \frac{25}{4} - \frac{10}{4} \)
\( 5d = \frac{15}{4} \)
\( d = \frac{15}{4 \times 5} \)
\( d = \frac{3}{4} \)
Substitute the value of \( d \) back into Equation 1:
\( a + \frac{3}{4} = \frac{5}{2} \)
\( a = \frac{5}{2} - \frac{3}{4} \)
\( a = \frac{10}{4} - \frac{3}{4} \)
\( a = \frac{7}{4} \)
Now we have the first term \( a = \frac{7}{4} \) and the common difference \( d = \frac{3}{4} \) for the A.P. The A.P. terms are \( \frac{7}{4}, \frac{7}{4} + \frac{3}{4}, \frac{7}{4} + 2(\frac{3}{4}), \dots \), which simplifies to \( \frac{7}{4}, \frac{10}{4}, \frac{13}{4}, \dots \).
The Harmonic Progression is found by taking the reciprocals of these A.P. terms.
Hence, the required H.P. is \( \frac{4}{7}, \frac{4}{10}, \frac{4}{13}, \dots \).
In simple words: Convert the H.P. terms into A.P. terms by taking their reciprocals. Then, set up equations for the first term (\( a \)) and common difference (\( d \)) of the A.P. Solve these equations to find \( a \) and \( d \). Finally, list the A.P. terms and take their reciprocals again to get the H.P.

🎯 Exam Tip: Always remember that if \( x \) is a term in H.P., then \( \frac{1}{x} \) is the corresponding term in A.P. This relationship is key to solving H.P. problems.

Question 4. If 7th term of H.P. is 17/2 and 11th term is 13/2 then find 20th term.
Answer:
If the 7th term of the Harmonic Progression (H.P.) is \( \frac{17}{2} \), then the 7th term of its corresponding Arithmetic Progression (A.P.) is the reciprocal, \( a + 6d = \frac{2}{17} \) (Equation 1).
If the 11th term of the H.P. is \( \frac{13}{2} \), then the 11th term of its corresponding A.P. is the reciprocal, \( a + 10d = \frac{2}{13} \) (Equation 2).
Now, we solve these two linear equations for \( a \) (first term of A.P.) and \( d \) (common difference of A.P.).
Subtract Equation 1 from Equation 2:
\( (a + 10d) - (a + 6d) = \frac{2}{13} - \frac{2}{17} \)
\( 4d = \frac{2 \times 17 - 2 \times 13}{13 \times 17} \)
\( 4d = \frac{34 - 26}{221} \)
\( 4d = \frac{8}{221} \)
\( d = \frac{8}{221 \times 4} \)
\( d = \frac{2}{221} \)
Substitute the value of \( d \) back into Equation 1:
\( a + 6 \left( \frac{2}{221} \right) = \frac{2}{17} \)
\( a + \frac{12}{221} = \frac{2}{17} \)
\( a = \frac{2}{17} - \frac{12}{221} \)
To subtract, find a common denominator, which is 221 (since \( 17 \times 13 = 221 \)).
\( a = \frac{2 \times 13}{17 \times 13} - \frac{12}{221} \)
\( a = \frac{26}{221} - \frac{12}{221} \)
\( a = \frac{14}{221} \)
Now we have \( a = \frac{14}{221} \) and \( d = \frac{2}{221} \). We need to find the 20th term of the A.P., \( T_{20} \).
\( T_{20} = a + 19d \)
\( T_{20} = \frac{14}{221} + 19 \left( \frac{2}{221} \right) \)
\( T_{20} = \frac{14 + 38}{221} \)
\( T_{20} = \frac{52}{221} \)
We can simplify this fraction. Both 52 and 221 are divisible by 13.
\( T_{20} = \frac{52 \div 13}{221 \div 13} = \frac{4}{17} \)
The 20th term of the corresponding H.P. is the reciprocal of the 20th term of the A.P.
Hence, the 20th term of the H.P. is \( \frac{17}{4} \).
In simple words: When given terms of an H.P., first convert them into A.P. terms by flipping them. Use these A.P. terms to find the first term (\( a \)) and common difference (\( d \)) of the A.P. Then, use \( a \) and \( d \) to find the specific A.P. term asked for. Finally, flip that A.P. term to get the answer for the H.P.

🎯 Exam Tip: Always simplify fractions at the end of your calculations. This makes the answer cleaner and often easier to verify.

Question 5. Find:
(i) 4 H.M. between 1 and 1/16.
(ii) 5 H.M. between 1/19 and 1/7.
(iii) 1 H.M. between \( -\frac{2}{5} \) and \( \frac{4}{25} \).
Answer:
(i) To find 4 Harmonic Means (H.M.) between 1 and \( \frac{1}{16} \), let the 4 H.M.s be \( H_1, H_2, H_3, H_4 \).
The sequence \( 1, H_1, H_2, H_3, H_4, \frac{1}{16} \) forms a Harmonic Progression.
The corresponding Arithmetic Progression (A.P.) will be the reciprocals of these terms: \( 1, \frac{1}{H_1}, \frac{1}{H_2}, \frac{1}{H_3}, \frac{1}{H_4}, 16 \).
For this A.P., the first term \( a = 1 \).
The last term is the 6th term (since there are 4 means plus 2 end terms). So, \( T_6 = 16 \).
Using the formula for the \( n^{th} \) term of an A.P., \( T_n = a + (n-1)d \):
\( 16 = 1 + (6-1)d \)
\( 16 = 1 + 5d \)
\( 15 = 5d \)
\( d = 3 \)
Now we can find the terms of the A.P. (which are the Arithmetic Means between 1 and 16):
\( A_1 = a + d = 1 + 3 = 4 \)
\( A_2 = a + 2d = 1 + 2(3) = 7 \)
\( A_3 = a + 3d = 1 + 3(3) = 10 \)
\( A_4 = a + 4d = 1 + 4(3) = 13 \)
The 4 Harmonic Means are the reciprocals of these Arithmetic Means.
Hence, the 4 H.M.s are \( \frac{1}{4}, \frac{1}{7}, \frac{1}{10}, \frac{1}{13} \).
(ii) To find 5 Harmonic Means (H.M.) between \( \frac{1}{19} \) and \( \frac{1}{7} \), let the 5 H.M.s be \( H_1, H_2, H_3, H_4, H_5 \).
The sequence \( \frac{1}{19}, H_1, H_2, H_3, H_4, H_5, \frac{1}{7} \) forms a Harmonic Progression.
The corresponding Arithmetic Progression (A.P.) will be the reciprocals of these terms: \( 19, \frac{1}{H_1}, \frac{1}{H_2}, \frac{1}{H_3}, \frac{1}{H_4}, \frac{1}{H_5}, 7 \).
For this A.P., the first term \( a = 19 \).
The last term is the 7th term (since there are 5 means plus 2 end terms). So, \( T_7 = 7 \).
Using the formula for the \( n^{th} \) term of an A.P., \( T_n = a + (n-1)d \):
\( 7 = 19 + (7-1)d \)
\( 7 = 19 + 6d \)
\( 7 - 19 = 6d \)
\( -12 = 6d \)
\( d = -2 \)
Now we can find the terms of the A.P. (which are the Arithmetic Means between 19 and 7):
\( A_1 = a + d = 19 + (-2) = 17 \)
\( A_2 = a + 2d = 19 + 2(-2) = 15 \)
\( A_3 = a + 3d = 19 + 3(-2) = 13 \)
\( A_4 = a + 4d = 19 + 4(-2) = 11 \)
\( A_5 = a + 5d = 19 + 5(-2) = 9 \)
The 5 Harmonic Means are the reciprocals of these Arithmetic Means.
Hence, the 5 H.M.s are \( \frac{1}{17}, \frac{1}{15}, \frac{1}{13}, \frac{1}{11}, \frac{1}{9} \).
(iii) To find 1 Harmonic Mean (H.M.) between \( -\frac{2}{5} \) and \( \frac{4}{25} \), let the H.M. be \( H \).
The sequence \( -\frac{2}{5}, H, \frac{4}{25} \) forms a Harmonic Progression.
The corresponding Arithmetic Progression (A.P.) will be the reciprocals of these terms: \( -\frac{5}{2}, \frac{1}{H}, \frac{25}{4} \).
In an A.P. of three terms, the middle term is the arithmetic mean of the first and third terms.
So, \( \frac{1}{H} = \frac{-\frac{5}{2} + \frac{25}{4}}{2} \)
First, find a common denominator for the terms in the numerator:
\( \frac{1}{H} = \frac{-\frac{10}{4} + \frac{25}{4}}{2} \)
\( \frac{1}{H} = \frac{\frac{15}{4}}{2} \)
\( \frac{1}{H} = \frac{15}{4 \times 2} \)
\( \frac{1}{H} = \frac{15}{8} \)
To find \( H \), take the reciprocal of \( \frac{15}{8} \).
Hence, the one H.M. is \( H = \frac{8}{15} \).
In simple words: To find Harmonic Means, turn the H.P. into an A.P. by taking reciprocals of the given numbers. Calculate the common difference of this A.P., then find the intermediate terms. These intermediate terms are Arithmetic Means. Finally, take the reciprocal of each Arithmetic Mean to get the Harmonic Means. For a single H.M., use the property that its reciprocal is the arithmetic mean of the reciprocals of the two end terms.

🎯 Exam Tip: Remember that for three numbers \( a, b, c \) in A.P., \( b = \frac{a+c}{2} \). For three numbers \( x, y, z \) in H.P., their reciprocals \( \frac{1}{x}, \frac{1}{y}, \frac{1}{z} \) are in A.P., so \( \frac{1}{y} = \frac{\frac{1}{x} + \frac{1}{z}}{2} \).

Question 6. If a, b, c are \( p^{th}, q^{th} \) and \( r^{th} \) terms of H.P. respectively then prove that \( bc(q - r) + ca(r - p) + ab(p - q) = 0 \).
Answer:
Given that a, b, c are the \( p^{th}, q^{th} \), and \( r^{th} \) terms of a Harmonic Progression (H.P.).
This means their reciprocals, \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \), are the \( p^{th}, q^{th} \), and \( r^{th} \) terms of a corresponding Arithmetic Progression (A.P.).
Let \( x \) be the first term and \( y \) be the common difference of this A.P.
Using the formula for the \( n^{th} \) term of an A.P., \( T_n = \text{first term} + (n-1) \times \text{common difference} \):
\( \frac{1}{a} = x + (p-1)y \) (Equation 1)
\( \frac{1}{b} = x + (q-1)y \) (Equation 2)
\( \frac{1}{c} = x + (r-1)y \) (Equation 3)
To prove the given identity, we will multiply Equation 1 by \( (q-r) \), Equation 2 by \( (r-p) \), and Equation 3 by \( (p-q) \), and then add the results.
Multiplying Equation 1 by \( (q-r) \):
\( \frac{1}{a}(q-r) = x(q-r) + (p-1)y(q-r) \)
Multiplying Equation 2 by \( (r-p) \):
\( \frac{1}{b}(r-p) = x(r-p) + (q-1)y(r-p) \)
Multiplying Equation 3 by \( (p-q) \):
\( \frac{1}{c}(p-q) = x(p-q) + (r-1)y(p-q) \)
Now, add these three new equations:
L.H.S.: \( \frac{1}{a}(q-r) + \frac{1}{b}(r-p) + \frac{1}{c}(p-q) \)
R.H.S.: \( x(q-r+r-p+p-q) + y[(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)] \)
Simplify the terms in the first bracket of the R.H.S.:
\( q-r+r-p+p-q = 0 \)
Simplify the terms in the second bracket of the R.H.S. by expanding them:
\( (pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q) \)
\( = pq - pr - q + r + qr - qp - r + p + rp - rq - p + q \)
\( = 0 \) (All terms cancel out)
So, the R.H.S. becomes \( x(0) + y(0) = 0 \).
Therefore, \( \frac{1}{a}(q-r) + \frac{1}{b}(r-p) + \frac{1}{c}(p-q) = 0 \).
Now, multiply the entire equation by \( abc \) to clear the denominators:
\( abc \left[ \frac{1}{a}(q-r) + \frac{1}{b}(r-p) + \frac{1}{c}(p-q) \right] = abc \times 0 \)
\( bc(q-r) + ac(r-p) + ab(p-q) = 0 \).
This proves the required identity. This property shows a cyclic relationship between terms in a harmonic progression.
In simple words: If you have three numbers in H.P. at certain positions, you can use their reciprocals to form an A.P. Then, by cleverly multiplying and adding the A.P. term formulas, all the terms involving the first term and common difference will cancel out, leaving zero. This leads directly to the equation we needed to prove.

🎯 Exam Tip: For proving identities involving H.P., always convert them to A.P. first. Algebraic manipulation of \( T_n \) equations for A.P. is a common technique to arrive at the desired result.

Question 7. If a, b, c are in H.P., then prove that a, a – c, a - b will be in H.P.
Answer:
Given that a, b, c are in Harmonic Progression (H.P.).
By the definition of H.P., their reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) must be in Arithmetic Progression (A.P.).
For three terms to be in A.P., the common difference between consecutive terms must be equal:
\( \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b} \)
\( \frac{a-b}{ab} = \frac{b-c}{bc} \)
Cross-multiplying, we get: \( c(a-b) = a(b-c) \) (Condition 1)

Now, we need to prove that a, \( a-c \), \( a-b \) are also in H.P.
For these terms to be in H.P., their reciprocals \( \frac{1}{a}, \frac{1}{a-c}, \frac{1}{a-b} \) must be in A.P.
This means the common difference between consecutive terms must be equal:
\( \frac{1}{a-c} - \frac{1}{a} = \frac{1}{a-b} - \frac{1}{a-c} \)
Let's simplify the Left Hand Side (L.H.S.):
\( \text{L.H.S.} = \frac{a - (a-c)}{a(a-c)} = \frac{a - a + c}{a(a-c)} = \frac{c}{a(a-c)} \)
Now, let's simplify the Right Hand Side (R.H.S.):
\( \text{R.H.S.} = \frac{(a-c) - (a-b)}{(a-b)(a-c)} = \frac{a - c - a + b}{(a-b)(a-c)} = \frac{b-c}{(a-b)(a-c)} \)
So, we need to prove: \( \frac{c}{a(a-c)} = \frac{b-c}{(a-b)(a-c)} \)
If \( a-c \neq 0 \), we can multiply both sides by \( (a-c) \):
\( \frac{c}{a} = \frac{b-c}{a-b} \)
Now, cross-multiply:
\( c(a-b) = a(b-c) \)
This equation is exactly Condition 1, which we derived from the fact that a, b, c are in H.P.
Since the condition holds true, it proves that \( \frac{1}{a}, \frac{1}{a-c}, \frac{1}{a-b} \) are in A.P.
Therefore, a, \( a-c \), \( a-b \) are in H.P. This shows an interesting property about terms derived from a given H.P.
In simple words: To prove that three new numbers are in H.P., we just need to show that their reciprocals form an A.P. We do this by checking if the difference between the second reciprocal and the first is the same as the difference between the third reciprocal and the second. If this matches a condition we already know is true (because the original numbers were in H.P.), then the proof is complete.

🎯 Exam Tip: When proving that a new set of terms forms an H.P., always aim to show that their reciprocals form an A.P. This is a standard and effective approach for such proofs.

Question 8. If a, b, c are in H.P., then prove that \( \frac{1}{b-a} + \frac{1}{b-c} = \frac{2}{b} \).
Answer:
Given that a, b, c are in Harmonic Progression (H.P.).
From the property of H.P., the middle term \( b \) can be expressed as: \( b = \frac{2ac}{a+c} \). (Equation 1)
Also, the reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in Arithmetic Progression (A.P.), which implies: \( \frac{1}{a} + \frac{1}{c} = \frac{2}{b} \). (Equation 2)

We need to prove that \( \frac{1}{b-a} + \frac{1}{b-c} = \frac{2}{b} \).
First, let's find expressions for \( (b-a) \) and \( (b-c) \) using Equation 1:
\( b-a = \frac{2ac}{a+c} - a \)
\( = \frac{2ac - a(a+c)}{a+c} \)
\( = \frac{2ac - a^2 - ac}{a+c} \)
\( = \frac{ac - a^2}{a+c} \)
\( = \frac{a(c-a)}{a+c} \)

Similarly for \( (b-c) \):
\( b-c = \frac{2ac}{a+c} - c \)
\( = \frac{2ac - c(a+c)}{a+c} \)
\( = \frac{2ac - ac - c^2}{a+c} \)
\( = \frac{ac - c^2}{a+c} \)
\( = \frac{c(a-c)}{a+c} \)

Now, let's substitute these expressions into the Left Hand Side (L.H.S.) of the equation we need to prove:
\( \text{L.H.S.} = \frac{1}{b-a} + \frac{1}{b-c} \)
\( = \frac{1}{\frac{a(c-a)}{a+c}} + \frac{1}{\frac{c(a-c)}{a+c}} \)
\( = \frac{a+c}{a(c-a)} + \frac{a+c}{c(a-c)} \)
To make the denominators consistent, we can write \( (a-c) \) as \( -(c-a) \):
\( = \frac{a+c}{a(c-a)} - \frac{a+c}{c(c-a)} \)
Now, take \( \frac{a+c}{c-a} \) as a common factor:
\( = \frac{a+c}{c-a} \left( \frac{1}{a} - \frac{1}{c} \right) \)
Combine the terms inside the parenthesis:
\( = \frac{a+c}{c-a} \left( \frac{c-a}{ac} \right) \)
The \( (c-a) \) terms cancel out:
\( = \frac{a+c}{ac} \)
Split this fraction into two terms:
\( = \frac{a}{ac} + \frac{c}{ac} \)
\( = \frac{1}{c} + \frac{1}{a} \)
From Equation 2, we know that \( \frac{1}{a} + \frac{1}{c} = \frac{2}{b} \).
So, \( \text{L.H.S.} = \frac{2}{b} \).
Thus, L.H.S. = R.H.S., which proves the identity. This identity shows a specific relationship between the terms of a harmonic progression.
In simple words: First, use the fact that if numbers are in H.P., then \( b \) can be written using \( a \) and \( c \). Then, use this to find out what \( b-a \) and \( b-c \) are. Put these into the left side of the equation you want to prove. Simplify everything, and you should find that it leads to the right side, which means the proof is done.

🎯 Exam Tip: When proving complex identities involving H.P., it's often helpful to first express \( b \) in terms of \( a \) and \( c \) using the H.P. definition, and then substitute these into the expression to simplify. Always keep the A.P. property \( \frac{1}{a} + \frac{1}{c} = \frac{2}{b} \) in mind for direct substitution.

Question 9. Find the Harmonic Mean of the roots of the quadratic equation \( ax^2 + bx + c = 0 \).
Answer:
Let the roots of the quadratic equation \( ax^2 + bx + c = 0 \) be \( \alpha \) and \( \beta \).
From the properties of quadratic equations, we know:
Sum of the roots: \( \alpha + \beta = -\frac{b}{a} \)
Product of the roots: \( \alpha\beta = \frac{c}{a} \)

The Harmonic Mean (H.M.) of two numbers \( \alpha \) and \( \beta \) is given by the formula:
\( H = \frac{2}{\frac{1}{\alpha} + \frac{1}{\beta}} \)
First, let's simplify the denominator \( \frac{1}{\alpha} + \frac{1}{\beta} \):
\( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta} \)
Now, substitute the sum and product of the roots into this expression:
\( \frac{\beta + \alpha}{\alpha\beta} = \frac{-\frac{b}{a}}{\frac{c}{a}} \)
\( = -\frac{b}{a} \times \frac{a}{c} \)
\( = -\frac{b}{c} \)
Now substitute this back into the H.M. formula:
\( H = \frac{2}{-\frac{b}{c}} \)
\( H = -\frac{2c}{b} \)
Hence, the Harmonic Mean of the roots of the quadratic equation \( ax^2 + bx + c = 0 \) is \( -\frac{2c}{b} \). This provides a quick way to find the harmonic mean directly from the coefficients of the quadratic equation.
In simple words: To find the Harmonic Mean of a quadratic equation's roots, first remember the formulas for the sum and product of roots. Then, use the H.M. formula \( \frac{2}{\frac{1}{\text{root1}} + \frac{1}{\text{root2}}} \). Replace \( \frac{1}{\text{root1}} + \frac{1}{\text{root2}} \) with \( \frac{\text{sum of roots}}{\text{product of roots}} \) and substitute the values from the quadratic equation. Simplify to get the final answer.

🎯 Exam Tip: Remember the relationships between the coefficients and roots of a quadratic equation (sum and product of roots). This is fundamental for solving problems involving means of roots.

Question 10. If \( p^{th} \) term of any H.P. is q and \( q^{th} \) term is p, then prove that \( (p + q)^{th} \) term will be \( \frac{pq}{(p+q)} \).
Answer:
Given that the \( p^{th} \) term of a Harmonic Progression (H.P.) is \( q \).
This means the \( p^{th} \) term of the corresponding Arithmetic Progression (A.P.) is \( \frac{1}{q} \).
Given that the \( q^{th} \) term of the H.P. is \( p \).
This means the \( q^{th} \) term of the corresponding A.P. is \( \frac{1}{p} \).

Let \( a \) be the first term and \( d \) be the common difference of the A.P.
Using the formula \( T_n = a + (n-1)d \):
For the \( p^{th} \) term: \( a + (p-1)d = \frac{1}{q} \) (Equation 1)
For the \( q^{th} \) term: \( a + (q-1)d = \frac{1}{p} \) (Equation 2)

Now, subtract Equation 2 from Equation 1 to find \( d \):
\( [a + (p-1)d] - [a + (q-1)d] = \frac{1}{q} - \frac{1}{p} \)
\( (p-1-q+1)d = \frac{p-q}{pq} \)
\( (p-q)d = \frac{p-q}{pq} \)
Assuming \( p \neq q \), we can divide both sides by \( (p-q) \):
\( d = \frac{1}{pq} \)

Substitute the value of \( d \) back into Equation 1 to find \( a \):
\( a + (p-1) \left( \frac{1}{pq} \right) = \frac{1}{q} \)
\( a = \frac{1}{q} - \frac{p-1}{pq} \)
To combine these terms, find a common denominator, which is \( pq \):
\( a = \frac{p}{pq} - \frac{p-1}{pq} \)
\( a = \frac{p - (p-1)}{pq} \)
\( a = \frac{p - p + 1}{pq} \)
\( a = \frac{1}{pq} \)

Now, we need to find the \( (p+q)^{th} \) term of the A.P. using \( a \) and \( d \):
\( T_{p+q} = a + (p+q-1)d \)
\( = \frac{1}{pq} + (p+q-1) \left( \frac{1}{pq} \right) \)
\( = \frac{1 + (p+q-1)}{pq} \)
\( = \frac{1 + p + q - 1}{pq} \)
\( = \frac{p+q}{pq} \)

Finally, the \( (p+q)^{th} \) term of the H.P. is the reciprocal of the \( (p+q)^{th} \) term of the A.P.
Hence, the \( (p+q)^{th} \) term of the H.P. is \( \frac{pq}{p+q} \). This is a classic result for such a specific H.P. condition.
In simple words: When given terms of an H.P. based on their position (like \( p^{th} \) term is \( q \)), convert them to A.P. terms by taking reciprocals. Set up two equations for the first term (\( a \)) and common difference (\( d \)) of the A.P. Solve these equations to find \( a \) and \( d \). Then, calculate the required term (here, the \( (p+q)^{th} \) term) of the A.P. and finally, take its reciprocal to get the answer for the H.P.

🎯 Exam Tip: This type of question frequently appears in exams. The key is to correctly set up the two simultaneous equations for \( a \) and \( d \) of the A.P. and solve them accurately.

Question 11. If roots of equation \( a(b - c)x^2 + b(c - a)x + c(a - b) = 0 \) are same then prove that a, b, c will be in H.P.
Answer:
Let the given quadratic equation be \( A x^2 + B x + C = 0 \), where:
\( A = a(b-c) \)
\( B = b(c-a) \)
\( C = c(a-b) \)

First, let's check if \( x=1 \) is a root of this equation by substituting \( x=1 \):
\( a(b-c)(1)^2 + b(c-a)(1) + c(a-b) = 0 \)
\( ab - ac + bc - ab + ca - cb = 0 \)
\( (ab - ab) + (-ac + ca) + (bc - cb) = 0 \)
\( 0 + 0 + 0 = 0 \)
Since substituting \( x=1 \) makes the equation true, \( x=1 \) is a root of the equation.

The problem states that the roots of the quadratic equation are *same*. Since one root is 1, the other root must also be 1.
For a quadratic equation \( A x^2 + B x + C = 0 \) with equal roots, we know that the sum of the roots is \( -\frac{B}{A} \) and the product of the roots is \( \frac{C}{A} \).
Since both roots are 1:
Product of roots \( = 1 \times 1 = 1 \).
So, \( \frac{C}{A} = 1 \).
Substitute the values of \( A \) and \( C \):
\( \frac{c(a-b)}{a(b-c)} = 1 \)
\( c(a-b) = a(b-c) \)
Expand both sides:
\( ac - bc = ab - ac \)
Rearrange the terms to group similar ones:
\( ac + ac = ab + bc \)
\( 2ac = ab + bc \)
To show that a, b, c are in H.P., we need to demonstrate that \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in A.P., which means \( \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \).
Divide the equation \( 2ac = ab + bc \) by \( abc \):
\( \frac{2ac}{abc} = \frac{ab}{abc} + \frac{bc}{abc} \)
\( \frac{2}{b} = \frac{1}{c} + \frac{1}{a} \)
This is the condition for a, b, c to be in H.P.
Hence, if the roots of the given quadratic equation are equal, then a, b, c will be in H.P. This shows an elegant connection between quadratic equations with repeated roots and harmonic progressions.
In simple words: First, check if \( x=1 \) is a root of the equation. If it is, and the problem says both roots are the same, then both roots must be 1. For a quadratic equation, the product of roots is \( C/A \). Set this to 1 and simplify the equation. You will find that it leads to the condition \( \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \), which is what it means for a, b, c to be in H.P.

🎯 Exam Tip: When a quadratic equation has coefficients that are expressions involving variables (like a, b, c here), always test for common roots like \( x=0 \) or \( x=1 \). If \( x=1 \) is a root and roots are equal, the solution becomes much simpler than using the discriminant formula \( B^2 - 4AC = 0 \).

Question 12. If a student, from his house goes to school at the speed of 8 km/h and returns with the speed of 6 km/h, then find its average speed where as distance between house and school is 6 km. Also verify the answer.
Answer:
When an object travels a certain distance at one speed and returns the same distance at another speed, the average speed is given by the Harmonic Mean (H.M.) of the two speeds. The actual distance (6 km) is not needed for the average speed calculation itself, but it is used for verification.
Given:
Speed from house to school (\( a \)) = 8 km/h
Speed from school to house (\( b \)) = 6 km/h

The formula for the Harmonic Mean (average speed for equal distances) is:
\( \text{Average Speed (H.M.)} = \frac{2ab}{a+b} \)
Substitute the given speeds:
\( \text{H.M.} = \frac{2 \times 8 \times 6}{8+6} \)
\( = \frac{96}{14} \)
\( = \frac{48}{7} \text{ km/h} \)
The average speed of the student is \( \frac{48}{7} \) km/h, which is approximately \( 6.86 \) km/h.

Verification:
Let the distance between the house and school be \( D = 6 \) km.
Total distance covered = Distance to school + Distance from school \( = D + D = 2D = 2 \times 6 = 12 \) km.

Time taken to go from house to school (\( t_1 \)) = \( \frac{\text{Distance}}{\text{Speed}} = \frac{6}{8} = \frac{3}{4} \) hours.
Time taken to return from school to house (\( t_2 \)) = \( \frac{\text{Distance}}{\text{Speed}} = \frac{6}{6} = 1 \) hour.

Total time taken = \( t_1 + t_2 = \frac{3}{4} + 1 = \frac{3}{4} + \frac{4}{4} = \frac{7}{4} \) hours.

Average speed = \( \frac{\text{Total Distance}}{\text{Total Time}} \)
\( = \frac{12}{\frac{7}{4}} \)
\( = \frac{12 \times 4}{7} \)
\( = \frac{48}{7} \text{ km/h} \)
The average speed calculated by the Harmonic Mean formula matches the average speed calculated by total distance divided by total time. This confirms the result. This principle is very useful in physics problems involving varying speeds over constant distances.
In simple words: When someone travels a distance at one speed and comes back the same distance at another speed, the average speed isn't just the simple average of the two speeds. Instead, you use a special formula called the Harmonic Mean: multiply the two speeds by 2, then divide by the sum of the speeds. To check, calculate the total distance traveled and the total time taken, then divide them. Both ways should give the same average speed.

🎯 Exam Tip: For problems involving average speed over equal distances, always use the Harmonic Mean formula \( \frac{2ab}{a+b} \). For unequal distances, you must calculate total distance divided by total time.