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

Question 1. Find three numbers in G.P. whose sum is 19 and product is 216.
Answer: Let the three numbers in G.P. be \( \frac{a}{r}, a, ar \).
Their product is given as 216:
\( \frac{a}{r} \times a \times ar = 216 \)
\( \implies a^3 = 216 \)
\( \implies a^3 = 6^3 \)
\( \implies a = 6 \)
Their sum is given as 19:
\( \frac{a}{r} + a + ar = 19 \)
Substitute \( a = 6 \) into the sum equation:
\( \frac{6}{r} + 6 + 6r = 19 \)
Subtract 6 from both sides:
\( \frac{6}{r} + 6r = 13 \)
Multiply the entire equation by \( r \):
\( 6 + 6r^2 = 13r \)
Rearrange into a quadratic equation:
\( 6r^2 - 13r + 6 = 0 \)
Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( r = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(6)(6)}}{2(6)} \)
\( r = \frac{13 \pm \sqrt{169 - 144}}{12} \)
\( r = \frac{13 \pm \sqrt{25}}{12} \)
\( r = \frac{13 \pm 5}{12} \)
We get two possible values for \( r \):
\( r_1 = \frac{13 + 5}{12} = \frac{18}{12} = \frac{3}{2} \)
\( r_2 = \frac{13 - 5}{12} = \frac{8}{12} = \frac{2}{3} \)
Case 1: When \( a = 6 \) and \( r = \frac{3}{2} \)
The numbers are: \( \frac{a}{r} = \frac{6}{3/2} = 6 \times \frac{2}{3} = 4 \)
\( a = 6 \)
\( ar = 6 \times \frac{3}{2} = 9 \)
The numbers are 4, 6, 9.
Case 2: When \( a = 6 \) and \( r = \frac{2}{3} \)
The numbers are: \( \frac{a}{r} = \frac{6}{2/3} = 6 \times \frac{3}{2} = 9 \)
\( a = 6 \)
\( ar = 6 \times \frac{2}{3} = 4 \)
The numbers are 9, 6, 4. Both sets of numbers are valid.
In simple words: We assume the three numbers in a G.P. are \( \frac{a}{r}, a, ar \). We use the given product to find the value of 'a'. Then, we use the given sum to find the value of 'r' by solving a quadratic equation. This gives two possible common ratios, leading to two sets of numbers for the G.P.

🎯 Exam Tip: When finding terms of a G.P. given their product, always assume the terms as \( \frac{a}{r}, a, ar \) to simplify calculations, as their product directly gives \( a^3 \). Remember to check both possible values of the common ratio \( r \).

Question 2. The sum of the first three terms of a G.P. is \( \frac { 13 }{ 12 } \) and their product is -1. Find the G.P.
Answer: Let the required G.P. be \( a, ar, ar^2 \).
Given that the product of the first three terms is -1:
\( a \times ar \times ar^2 = -1 \)
\( \implies a^3r^3 = -1 \)
\( \implies (ar)^3 = -1 \)
\( \implies ar = -1 \)
From this, we can express \( a \) in terms of \( r \):
\( a = -\frac{1}{r} \) ... (1)
Given that the sum of the first three terms is \( \frac{13}{12} \):
\( a + ar + ar^2 = \frac{13}{12} \)
Substitute \( ar = -1 \) and \( a = -\frac{1}{r} \) into the sum equation:
\( -\frac{1}{r} + (-1) + (-1)r = \frac{13}{12} \)
\( -\frac{1}{r} - 1 - r = \frac{13}{12} \)
Multiply the entire equation by \( 12r \) to clear denominators:
\( -12 - 12r - 12r^2 = 13r \)
Move all terms to one side to form a quadratic equation:
\( 12r^2 + 12r + 13r + 12 = 0 \)
\( \implies 12r^2 + 25r + 12 = 0 \)
Now, solve this quadratic equation for \( r \) using the quadratic formula:
\( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( r = \frac{-25 \pm \sqrt{25^2 - 4(12)(12)}}{2(12)} \)
\( r = \frac{-25 \pm \sqrt{625 - 576}}{24} \)
\( r = \frac{-25 \pm \sqrt{49}}{24} \)
\( r = \frac{-25 \pm 7}{24} \)
We get two possible values for \( r \):
\( r_1 = \frac{-25 + 7}{24} = \frac{-18}{24} = -\frac{3}{4} \)
\( r_2 = \frac{-25 - 7}{24} = \frac{-32}{24} = -\frac{4}{3} \)
Case 1: When \( r = -\frac{3}{4} \)
From \( a = -\frac{1}{r} \): \( a = -\frac{1}{(-3/4)} = \frac{4}{3} \)
The G.P. terms are: \( a = \frac{4}{3} \)
\( ar = \frac{4}{3} \times (-\frac{3}{4}) = -1 \)
\( ar^2 = \frac{4}{3} \times (-\frac{3}{4})^2 = \frac{4}{3} \times \frac{9}{16} = \frac{3}{4} \)
So, the G.P. is \( \frac{4}{3}, -1, \frac{3}{4}, \dots \)
Case 2: When \( r = -\frac{4}{3} \)
From \( a = -\frac{1}{r} \): \( a = -\frac{1}{(-4/3)} = \frac{3}{4} \)
The G.P. terms are: \( a = \frac{3}{4} \)
\( ar = \frac{3}{4} \times (-\frac{4}{3}) = -1 \)
\( ar^2 = \frac{3}{4} \times (-\frac{4}{3})^2 = \frac{3}{4} \times \frac{16}{9} = \frac{4}{3} \)
So, the G.P. is \( \frac{3}{4}, -1, \frac{4}{3}, \dots \)
In simple words: We find the first term 'a' and common ratio 'r' of the G.P. by using the given product and sum of the terms. We solve a cubic equation for the product and a quadratic equation for the sum to find 'a' and 'r'. Since there are two possible values for 'r', we get two possible G.P. sequences.

🎯 Exam Tip: Always remember that the product of terms in a G.P. can help you find the middle term directly, which simplifies the process of finding 'a'. Be careful with signs when dealing with negative common ratios.

Question 3. The product of first three terms of a G.P. is 1000. If we add 6 to its second term and 7 to its third term, the resulting three terms form an A.P. Find the terms of the G.P.
Answer: Let the required G.P. be \( a, ar, ar^2 \).
Given that the product of the first three terms is 1000:
\( a \times ar \times ar^2 = 1000 \)
\( \implies a^3r^3 = 1000 \)
\( \implies (ar)^3 = 10^3 \)
\( \implies ar = 10 \)
From this, we can express \( a \) in terms of \( r \):
\( a = \frac{10}{r} \) ... (1)
The original G.P. terms are \( \frac{10}{r}, 10, 10r \).
If we add 6 to the second term and 7 to the third term, the new terms are:
First term: \( a = \frac{10}{r} \)
Second term: \( ar + 6 = 10 + 6 = 16 \)
Third term: \( ar^2 + 7 = 10r + 7 \)
These new terms form an A.P. For three terms to be in A.P., the middle term is the average of the first and third terms, or \( 2 \times (\text{middle term}) = (\text{first term}) + (\text{third term}) \).
\( 2(16) = \frac{10}{r} + (10r + 7) \)
\( 32 = \frac{10}{r} + 10r + 7 \)
Subtract 7 from both sides:
\( 25 = \frac{10}{r} + 10r \)
Multiply the entire equation by \( r \) to clear the denominator:
\( 25r = 10 + 10r^2 \)
Rearrange into a quadratic equation:
\( 10r^2 - 25r + 10 = 0 \)
Divide the entire equation by 5 to simplify:
\( 2r^2 - 5r + 2 = 0 \)
Factor the quadratic equation:
\( 2r^2 - 4r - r + 2 = 0 \)
\( 2r(r - 2) - 1(r - 2) = 0 \)
\( (r - 2)(2r - 1) = 0 \)
This gives two possible values for \( r \):
\( r = 2 \) or \( r = \frac{1}{2} \)
Case 1: When \( r = 2 \)
From (1), \( a = \frac{10}{r} = \frac{10}{2} = 5 \)
The terms of the G.P. are \( a, ar, ar^2 \):
\( 5, 5 \times 2, 5 \times 2^2 \)
\( 5, 10, 20 \)
Case 2: When \( r = \frac{1}{2} \)
From (1), \( a = \frac{10}{r} = \frac{10}{1/2} = 20 \)
The terms of the G.P. are \( a, ar, ar^2 \):
\( 20, 20 \times \frac{1}{2}, 20 \times (\frac{1}{2})^2 \)
\( 20, 10, 5 \)
Both sets of terms are valid G.P.s.
In simple words: First, we use the product of the three G.P. terms to find a relationship between the first term 'a' and the common ratio 'r'. Then, we use the condition that after adding to the second and third terms, they form an A.P. This helps us create an equation to find 'r'. We solve for 'r', then find 'a', and list the G.P. terms.

🎯 Exam Tip: When a problem involves both A.P. and G.P. conditions, solve for common elements like 'a' and 'r' step-by-step. Remember the property for three terms in A.P.: \( 2b = a + c \).

Question 4. If a, b, c are in G.P show that the following are also in G.P
(i) \( \frac { 1 }{ a }, \frac { 1 }{ b }, \frac { 1 }{ c } \)
(ii) \( \frac{1}{a^2}, \frac{1}{b^2}, \frac{1}{c^2} \)
(iii) \( a^2, b^2, c^2 \)
(iv) \( b^2c^2, c^2a^2, a^2b^2 \)
Answer: Given that a, b, c are in G.P. This means the square of the middle term is equal to the product of the other two terms:
\( b^2 = ac \) ... (1)

(i) To show that \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in G.P., we need to prove that the square of the middle term equals the product of the other two.
We need to show \( (\frac{1}{b})^2 = (\frac{1}{a}) \times (\frac{1}{c}) \).
From (1), we have \( b^2 = ac \).
Taking the reciprocal of both sides: \( \frac{1}{b^2} = \frac{1}{ac} \)
This can be written as: \( (\frac{1}{b})^2 = \frac{1}{a} \times \frac{1}{c} \)
Since this condition is met, \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are also in G.P. This shows that the reciprocals of terms in a G.P. also form a G.P.

(ii) To show that \( \frac{1}{a^2}, \frac{1}{b^2}, \frac{1}{c^2} \) are in G.P., we need to prove that \( (\frac{1}{b^2})^2 = (\frac{1}{a^2}) \times (\frac{1}{c^2}) \).
From (1), \( b^2 = ac \).
Squaring both sides of (1): \( (b^2)^2 = (ac)^2 \)
\( \implies b^4 = a^2c^2 \)
Taking the reciprocal of both sides: \( \frac{1}{b^4} = \frac{1}{a^2c^2} \)
This can be written as: \( (\frac{1}{b^2})^2 = \frac{1}{a^2} \times \frac{1}{c^2} \)
Since this condition is met, \( \frac{1}{a^2}, \frac{1}{b^2}, \frac{1}{c^2} \) are also in G.P.

(iii) To show that \( a^2, b^2, c^2 \) are in G.P., we need to prove that \( (b^2)^2 = a^2 \times c^2 \).
From (1), \( b^2 = ac \).
Squaring both sides of (1): \( (b^2)^2 = (ac)^2 \)
\( \implies b^4 = a^2c^2 \)
This is exactly the condition we need: \( (b^2)^2 = a^2 \times c^2 \).
Thus, \( a^2, b^2, c^2 \) are also in G.P.

(iv) To show that \( b^2c^2, c^2a^2, a^2b^2 \) are in G.P., we need to prove that \( (c^2a^2)^2 = (b^2c^2) \times (a^2b^2) \).
Let's simplify the right-hand side (R.H.S.):
R.H.S. \( = (b^2c^2) \times (a^2b^2) = a^2 b^4 c^2 \)
From (1), \( b^2 = ac \). This means \( b^4 = (ac)^2 = a^2c^2 \).
Substitute \( b^4 = a^2c^2 \) into the R.H.S.:
R.H.S. \( = a^2 (a^2c^2) c^2 = a^4 c^4 \)
Now, let's look at the left-hand side (L.H.S.):
L.H.S. \( = (c^2a^2)^2 = c^4 a^4 \)
Since L.H.S. = R.H.S. \( (a^4c^4 = a^4c^4) \), the condition is met.
Thus, \( b^2c^2, c^2a^2, a^2b^2 \) are also in G.P.
In simple words: If three numbers 'a', 'b', 'c' are in a Geometric Progression (G.P.), it means the square of the middle number 'b' is equal to the product of the first and last numbers, \( b^2 = ac \). We use this main rule to prove that other sets of numbers formed from 'a', 'b', 'c' also follow the G.P. rule by checking if the square of their middle term equals the product of their first and last terms. For example, squaring the terms of a G.P. still results in a G.P.

🎯 Exam Tip: The core property for terms in a G.P. is that the ratio of consecutive terms is constant (common ratio) or, for three terms, \( b^2 = ac \). Always use this fundamental property to prove new sequences are G.P.s.

Question 5. If a, b, c, d are in G.P., show that the following are also in G.P.
(i) \( a + b, b + c, c + d \)
(ii) \( a^2 + b^2, b^2 + c^2, c^2 + d^2 \)
Answer: Given that a, b, c, d are in G.P. Let 'r' be the common ratio.
Then, we can write the terms as:
\( b = ar \)
\( c = br = ar^2 \)
\( d = cr = ar^3 \)

(i) To show that \( a + b, b + c, c + d \) are in G.P., we need to prove that \( (b+c)^2 = (a+b)(c+d) \).
Let's evaluate the Left Hand Side (L.H.S.):
L.H.S. \( = (b+c)^2 \)
Substitute \( b=ar \) and \( c=ar^2 \):
\( = (ar + ar^2)^2 \)
\( = [ar(1+r)]^2 \)
\( = a^2r^2(1+r)^2 \)
Now, let's evaluate the Right Hand Side (R.H.S.):
R.H.S. \( = (a+b)(c+d) \)
Substitute \( b=ar, c=ar^2, d=ar^3 \):
\( = (a+ar)(ar^2+ar^3) \)
Factor out common terms:
\( = a(1+r) \cdot ar^2(1+r) \)
\( = a^2r^2(1+r)^2 \)
Since L.H.S. = R.H.S., the condition for a G.P. is met.
Therefore, \( a + b, b + c, c + d \) are in G.P.

(ii) To show that \( a^2 + b^2, b^2 + c^2, c^2 + d^2 \) are in G.P., we need to prove that \( (b^2+c^2)^2 = (a^2+b^2)(c^2+d^2) \).
Let's evaluate the Left Hand Side (L.H.S.):
L.H.S. \( = (b^2+c^2)^2 \)
Substitute \( b=ar \) and \( c=ar^2 \):
\( = ((ar)^2 + (ar^2)^2)^2 \)
\( = (a^2r^2 + a^2r^4)^2 \)
Factor out \( a^2r^2 \):
\( = [a^2r^2(1+r^2)]^2 \)
\( = a^4r^4(1+r^2)^2 \)
Now, let's evaluate the Right Hand Side (R.H.S.):
R.H.S. \( = (a^2+b^2)(c^2+d^2) \)
Substitute \( b=ar, c=ar^2, d=ar^3 \):
\( = (a^2 + (ar)^2)((ar^2)^2 + (ar^3)^2) \)
\( = (a^2 + a^2r^2)(a^2r^4 + a^2r^6) \)
Factor out common terms:
\( = a^2(1+r^2) \cdot a^2r^4(1+r^2) \)
\( = a^4r^4(1+r^2)^2 \)
Since L.H.S. = R.H.S., the condition for a G.P. is met.
Therefore, \( a^2 + b^2, b^2 + c^2, c^2 + d^2 \) are in G.P.
In simple words: When four numbers are in a G.P., we can write them using 'a' (first term) and 'r' (common ratio). To show that new sets of numbers also form a G.P., we check if the square of the middle term equals the product of the first and last terms for those new sets. We substitute the 'ar' forms of the terms into these equations and simplify to see if both sides are equal. This confirms the new sequences are also G.P.s.

🎯 Exam Tip: When working with four or more terms in a G.P., expressing them in terms of 'a' and 'r' is crucial. This makes it easier to manipulate algebraic expressions and prove the required properties by checking if \( B^2=AC \).

Question 6. If a, b, c, d are in G.P., prove that
(i) \( (b + c)(b + d) = (c + a)(c + d) \)
(ii) \( (a – d)^2 = (b − c)^2 + (c − a)^2 + (d – b)^2 \)
Answer: Given that a, b, c, d are in G.P. Let 'r' be the common ratio. Since \( r \neq 0 \), we can write the terms as:
\( b = ar \)
\( c = br = ar^2 \)
\( d = cr = ar^3 \)

(i) To prove: \( (b + c)(b + d) = (c + a)(c + d) \)
Let's evaluate the Left Hand Side (L.H.S.):
L.H.S. \( = (b + c)(b + d) \)
Substitute \( b=ar, c=ar^2, d=ar^3 \):
\( = (ar + ar^2)(ar + ar^3) \)
Factor out common terms:
\( = ar(1 + r) \cdot ar(1 + r^2) \)
\( = a^2r^2(1 + r)(1 + r^2) \)
Now, let's evaluate the Right Hand Side (R.H.S.):
R.H.S. \( = (c + a)(c + d) \)
Substitute \( a, c=ar^2, d=ar^3 \):
\( = (ar^2 + a)(ar^2 + ar^3) \)
Factor out common terms:
\( = a(r^2 + 1) \cdot ar^2(1 + r) \)
\( = a^2r^2(1 + r^2)(1 + r) \)
Since L.H.S. = R.H.S., the identity is proven.

(ii) To prove: \( (a – d)^2 = (b − c)^2 + (c − a)^2 + (d – b)^2 \)
Let's evaluate the Left Hand Side (L.H.S.):
L.H.S. \( = (a – d)^2 \)
Substitute \( d=ar^3 \):
\( = (a - ar^3)^2 \)
Factor out 'a':
\( = [a(1 - r^3)]^2 \)
\( = a^2(1 - r^3)^2 \)
Now, let's evaluate the Right Hand Side (R.H.S.):
R.H.S. \( = (b − c)^2 + (c − a)^2 + (d – b)^2 \)
Substitute \( b=ar, c=ar^2, d=ar^3 \):
\( = (ar - ar^2)^2 + (ar^2 - a)^2 + (ar^3 - ar)^2 \)
Factor out common terms from each parenthesis:
\( = [ar(1 - r)]^2 + [a(r^2 - 1)]^2 + [ar(r^2 - 1)]^2 \)
\( = a^2r^2(1 - r)^2 + a^2(r^2 - 1)^2 + a^2r^2(r^2 - 1)^2 \)
Factor out \( a^2 \):
\( = a^2 [r^2(1 - r)^2 + (r^2 - 1)^2 + r^2(r^2 - 1)^2] \)
Now, notice that \( (r^2 - 1) = (r - 1)(r + 1) \). So \( (r^2 - 1)^2 = (r - 1)^2(r + 1)^2 \).
Also, \( (1 - r)^2 = (r - 1)^2 \).
Substitute these into the expression:
\( = a^2 [r^2(r - 1)^2 + (r - 1)^2(r + 1)^2 + r^2(r - 1)^2(r + 1)^2] \)
Factor out \( (r - 1)^2 \):
\( = a^2(r - 1)^2 [r^2 + (r + 1)^2 + r^2(r + 1)^2] \)
Expand \( (r + 1)^2 = r^2 + 2r + 1 \):
\( = a^2(r - 1)^2 [r^2 + (r^2 + 2r + 1) + r^2(r^2 + 2r + 1)] \)
\( = a^2(r - 1)^2 [r^2 + r^2 + 2r + 1 + r^4 + 2r^3 + r^2] \)
\( = a^2(r - 1)^2 [r^4 + 2r^3 + 3r^2 + 2r + 1] \)
This expression \( (r^4 + 2r^3 + 3r^2 + 2r + 1) \) is actually the expansion of \( (r^2+r+1)^2 \).
So, \( = a^2(r - 1)^2 (r^2 + r + 1)^2 \)
\( = a^2 [(r - 1)(r^2 + r + 1)]^2 \)
Recall the algebraic identity: \( (x - y)(x^2 + xy + y^2) = x^3 - y^3 \).
Applying this, \( (r - 1)(r^2 + r + 1) = r^3 - 1^3 = r^3 - 1 \).
So, R.H.S. \( = a^2 (r^3 - 1)^2 \)
\( = a^2 (1 - r^3)^2 \)
Since L.H.S. \( = a^2(1 - r^3)^2 \), we have L.H.S. = R.H.S.
Thus, the identity is proven.
In simple words: When a, b, c, d are in a Geometric Progression (G.P.), we can write each term using 'a' and 'r' (the common ratio). For part (i), we put these 'ar' forms into both sides of the equation and show that they become equal. For part (ii), it's a longer process: we calculate each side of the equation separately, substitute the 'ar' forms, and simplify using algebraic identities like \( (x-y)(x^2+xy+y^2)=x^3-y^3 \) to show that both sides ultimately result in the same expression.

🎯 Exam Tip: For proving identities involving terms of a G.P., it's best to express all terms as \( a, ar, ar^2, ar^3, \dots \) and substitute them into the L.H.S. and R.H.S. separately. Then simplify each side to show they are equal. Algebraic identities like \( (x^3 - y^3) \) are very useful.

Question 7. If the pth, qth and rth terms of an A.P. are in G.P., prove that the common ratio of the G.P. is \( \frac{q-r}{p-q} \).
Answer: Let 'A' be the first term and 'D' be the common difference of the A.P.
The pth term of the A.P. is \( T_p = A + (p-1)D \).
The qth term of the A.P. is \( T_q = A + (q-1)D \).
The rth term of the A.P. is \( T_r = A + (r-1)D \).
Given that \( T_p, T_q, T_r \) are in G.P. This means the common ratio of the G.P. can be found by dividing consecutive terms:
Common ratio \( = \frac{T_q}{T_p} = \frac{T_r}{T_q} \)
So, \( \frac{A + (q-1)D}{A + (p-1)D} = \frac{A + (r-1)D}{A + (q-1)D} \)
Let \( k \) be the common ratio of the G.P.
So, \( \frac{T_q}{T_p} = k \) and \( \frac{T_r}{T_q} = k \).
From a property of proportions, if \( \frac{X}{Y} = \frac{Z}{W} \), then \( \frac{X-Z}{Y-W} \) is also equal to that ratio, provided \( Y \neq W \).
Applying this property to \( \frac{A + (q-1)D}{A + (p-1)D} = \frac{A + (r-1)D}{A + (q-1)D} \):
The common ratio \( k = \frac{(A + (q-1)D) - (A + (r-1)D)}{(A + (p-1)D) - (A + (q-1)D)} \)
\( \implies k = \frac{A + qD - D - A - rD + D}{A + pD - D - A - qD + D} \)
\( \implies k = \frac{(q-r)D}{(p-q)D} \)
If \( D \neq 0 \) and \( p \neq q \), then we can cancel \( D \):
\( \implies k = \frac{q-r}{p-q} \)
Hence, the common ratio of the G.P. is \( \frac{q-r}{p-q} \). This property helps find the ratio of terms in an A.P. when they form a G.P.
In simple words: We start with the formulas for the pth, qth, and rth terms of an Arithmetic Progression (A.P.). Since these terms also form a Geometric Progression (G.P.), their common ratio can be found by dividing any term by the one before it. We set up an equation using this G.P. common ratio property and simplify it. This simplification shows that the G.P. common ratio is equal to \( \frac{q-r}{p-q} \).

🎯 Exam Tip: When dealing with terms of an A.P. that are also in a G.P., remember the formula for the nth term of an A.P. \( T_n = A + (n-1)D \). The key property for G.P. is the constant ratio of consecutive terms. Using properties of ratios \( \frac{a}{b} = \frac{c}{d} \implies \frac{a-c}{b-d} = \frac{a}{b} \) is a neat trick here.

Question 8. If \( \frac{1}{x+y}, \frac{1}{2 y}, \frac{1}{y+z} \) are the three consecutive terms of an A.P. prove that x, y, z are the three consecutive terms of a G.P.
Answer: Given that \( \frac{1}{x+y}, \frac{1}{2y}, \frac{1}{y+z} \) are in A.P.
For three terms \( A, B, C \) to be in A.P., the middle term is the average of the first and third terms, which means \( 2B = A + C \).
So, \( 2 \left(\frac{1}{2y}\right) = \frac{1}{x+y} + \frac{1}{y+z} \)
\( \implies \frac{1}{y} = \frac{1(y+z) + 1(x+y)}{(x+y)(y+z)} \)
\( \implies \frac{1}{y} = \frac{y+z+x+y}{(x+y)(y+z)} \)
\( \implies \frac{1}{y} = \frac{x+2y+z}{(x+y)(y+z)} \)
Cross-multiply the terms:
\( (x+y)(y+z) = y(x+2y+z) \)
Expand both sides of the equation:
\( xy + xz + y^2 + yz = xy + 2y^2 + yz \)
Subtract \( xy \) from both sides:
\( xz + y^2 + yz = 2y^2 + yz \)
Subtract \( yz \) from both sides:
\( xz + y^2 = 2y^2 \)
Subtract \( y^2 \) from both sides:
\( xz = y^2 \)
This condition, \( y^2 = xz \), is the definition for three terms \( x, y, z \) to be in G.P.
Therefore, x, y, z are in G.P.
In simple words: We are given that three fractions are in an Arithmetic Progression (A.P.). We use the rule for A.P. (the middle term is the average of the other two) to set up an equation. By simplifying this equation step-by-step, we find that \( y^2 = xz \). This exact equation is the rule that defines numbers in a Geometric Progression (G.P.). So, we prove that x, y, and z must be in G.P.

🎯 Exam Tip: Remember the fundamental definitions: For A.P., \( 2B = A+C \), and for G.P., \( B^2 = AC \). When asked to prove a relationship, start with the given information (A.P. in this case), use its definition to form an equation, and then algebraically manipulate it to arrive at the definition of the desired relationship (G.P.).