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

Question 1. Find :
(i) the 7th term of 2, 4, 8, .....
(ii) the 9th term of 1, \( \frac { 1 }{ 2 } \), \( \frac{1}{2^2} \), ....
(iii) the nth term of \( \frac { 15 }{ 8 } \), \( \frac { 3 }{ 8 } \), \( \frac { 3 }{ 40 } \), ....
Answer:
(i) The given sequence is 2, 4, 8, ... This is a Geometric Progression (G.P.). The first term \( a = 2 \). The common ratio \( r \) is found by dividing a term by its previous term, so \( r = \frac{4}{2} = 2 \). To find the 7th term \( T_7 \), we use the G.P. formula \( T_n = ar^{n-1} \):
\( T_7 = 2 \times 2^{7-1} \)
\( T_7 = 2 \times 2^6 \)
\( T_7 = 2 \times 64 \)
\( T_7 = 128 \)
(ii) The given sequence is 1, \( \frac{1}{2} \), \( \frac{1}{2^2} \), ... This is a G.P. The first term \( a = 1 \). The common ratio \( r = \frac{1/2}{1} = \frac{1}{2} \). To find the 9th term \( T_9 \), we use the G.P. formula \( T_n = ar^{n-1} \):
\( T_9 = 1 \times \left(\frac{1}{2}\right)^{9-1} \)
\( T_9 = 1 \times \left(\frac{1}{2}\right)^8 \)
\( T_9 = \frac{1}{2^8} \)
\( T_9 = \frac{1}{256} \)
(iii) The given sequence is \( \frac{15}{8} \), \( \frac{3}{8} \), \( \frac{3}{40} \), ... This is a G.P. The first term \( a = \frac{15}{8} \). The common ratio \( r = \frac{3/8}{15/8} = \frac{3}{15} = \frac{1}{5} \). To find the nth term \( T_n \), we use the G.P. formula \( T_n = ar^{n-1} \):
\( T_n = \frac{15}{8} \times \left(\frac{1}{5}\right)^{n-1} \)
In simple words: For each sequence, first find the starting number and the multiplication factor. Then, use the general formula to find the specific term asked for. Remember, the common ratio shows how each number in the sequence is related to the previous one.

🎯 Exam Tip: Always identify if a sequence is an Arithmetic Progression (A.P.) or a Geometric Progression (G.P.) first, as their formulas for \( n^{th} \) term and sum are different.

Question 2. The second term of a G.P. is 18 and the fifth term is 486. Find :
(i) the first term,
(ii) the common ratio
Answer:
Let the first term of the G.P. be \( a \) and the common ratio be \( r \).
The formula for the \( n^{th} \) term of a G.P. is \( T_n = ar^{n-1} \).
We are given that the second term \( T_2 = 18 \):
\( ar^{2-1} = 18 \)
\( \implies ar = 18 \) (Equation 1)
We are also given that the fifth term \( T_5 = 486 \):
\( ar^{5-1} = 486 \)
\( \implies ar^4 = 486 \) (Equation 2)
To find the common ratio \( r \), divide Equation 2 by Equation 1:
\( \frac{ar^4}{ar} = \frac{486}{18} \)
\( \implies r^3 = 27 \)
\( \implies r^3 = 3^3 \)
\( \implies r = 3 \)
So, the common ratio is 3.
Now, substitute \( r=3 \) into Equation 1 to find the first term \( a \):
\( a \times 3 = 18 \)
\( \implies a = \frac{18}{3} \)
\( \implies a = 6 \)
So, the first term is 6.
In simple words: We used the given terms to set up two equations. By dividing these equations, we found the common multiplication factor (ratio) for the pattern, which is 3. Then, we used this ratio to find the very first number in the pattern, which is 6.

🎯 Exam Tip: When given two non-consecutive terms of a G.P., dividing the equations formed from the general term formula is an efficient way to find the common ratio.

Question 3. Find the value of x for which x + 9, x - 6, 4 are the first three terms of a geometrical progression and calculate the fourth term of progression in each case.
Answer:
Given that \( x+9 \), \( x-6 \), and \( 4 \) are the first three terms of a Geometric Progression (G.P.).
In a G.P., the square of the middle term is equal to the product of the first and third terms. If \( a, b, c \) are in G.P., then \( b^2 = ac \).
So, we can write the equation:
\( (x-6)^2 = (x+9) \times 4 \)
Expand both sides:
\( x^2 - 12x + 36 = 4x + 36 \)
Rearrange the terms to form a quadratic equation:
\( x^2 - 12x - 4x + 36 - 36 = 0 \)
\( x^2 - 16x = 0 \)
Factor out \( x \):
\( x(x - 16) = 0 \)
This gives two possible values for \( x \):
\( x = 0 \) or \( x - 16 = 0 \implies x = 16 \)
We need to calculate the fourth term for each case:
**Case 1: When \( x = 0 \)**
The three terms of the G.P. are:
First term: \( x+9 = 0+9 = 9 \)
Second term: \( x-6 = 0-6 = -6 \)
Third term: \( 4 \)
So the G.P. is \( 9, -6, 4, ... \).
The first term \( a = 9 \).
The common ratio \( r = \frac{-6}{9} = -\frac{2}{3} \).
The fourth term \( T_4 = ar^3 = 9 \times \left(-\frac{2}{3}\right)^3 \)
\( T_4 = 9 \times \left(-\frac{8}{27}\right) \)
\( T_4 = -\frac{72}{27} \)
\( T_4 = -\frac{8}{3} \)
**Case 2: When \( x = 16 \)**
The three terms of the G.P. are:
First term: \( x+9 = 16+9 = 25 \)
Second term: \( x-6 = 16-6 = 10 \)
Third term: \( 4 \)
So the G.P. is \( 25, 10, 4, ... \).
The first term \( a = 25 \).
The common ratio \( r = \frac{10}{25} = \frac{2}{5} \).
The fourth term \( T_4 = ar^3 = 25 \times \left(\frac{2}{5}\right)^3 \)
\( T_4 = 25 \times \frac{8}{125} \)
\( T_4 = \frac{25 \times 8}{125} \)
\( T_4 = \frac{200}{125} \)
\( T_4 = \frac{8}{5} \)
In simple words: We used the rule that in a multiplying pattern, the middle number squared is equal to the first times the third. This helped us find two possible values for 'x'. For each 'x' value, we then listed the numbers and calculated what the next number in that pattern would be.

🎯 Exam Tip: Remember that quadratic equations often yield multiple solutions, and each solution for 'x' must be evaluated fully to determine all possible sequences and their terms.

Question 4. If 5, x, y, z 405 are the first five terms of a geometric progression, find the values of x, y and z.
Answer:
Given that 5, x, y, z, 405 are the first five terms of a Geometric Progression (G.P.).
Here, the first term \( a = 5 \).
The fifth term \( T_5 = 405 \).
The formula for the \( n^{th} \) term of a G.P. is \( T_n = ar^{n-1} \). So, for the fifth term:
\( T_5 = ar^{5-1} = ar^4 \)
Substitute the known values:
\( 405 = 5 \times r^4 \)
Divide both sides by 5:
\( r^4 = \frac{405}{5} \)
\( r^4 = 81 \)
We know that \( 3^4 = 81 \), so:
\( r^4 = 3^4 \)
\( \implies r = 3 \)
Now that we have the common ratio \( r=3 \), we can find \( x, y, \) and \( z \):
\( x = T_2 = ar = 5 \times 3 = 15 \)
\( y = T_3 = ar^2 = 5 \times 3^2 = 5 \times 9 = 45 \)
\( z = T_4 = ar^3 = 5 \times 3^3 = 5 \times 27 = 135 \)
Thus, the values are \( x=15 \), \( y=45 \), and \( z=135 \).
In simple words: We have a multiplying pattern where the first number is 5 and the fifth is 405. We found the hidden multiplication factor (common ratio) is 3. Then, we used this factor to fill in the missing numbers: x is 15, y is 45, and z is 135.

🎯 Exam Tip: Remember that \( r^4 = 81 \) could also mean \( r = -3 \). However, if not specified, positive values are generally assumed for a geometric progression's common ratio in such contexts, especially if terms are implicitly increasing.

Question 5. Insert 3 geometric means between 16 and 256.
Answer:
To insert 3 geometric means (GM) between 16 and 256 means to form a Geometric Progression (G.P.) where 16 is the first term and 256 is the fifth term, with three terms in between them.
Let the G.P. be \( 16, G_1, G_2, G_3, 256 \).
Here, the first term \( a = 16 \).
The fifth term \( T_5 = 256 \).
Using the G.P. formula \( T_n = ar^{n-1} \):
\( T_5 = ar^{5-1} = ar^4 \)
Substitute the known values:
\( 256 = 16 \times r^4 \)
Divide by 16:
\( r^4 = \frac{256}{16} \)
\( r^4 = 16 \)
We know that \( 2^4 = 16 \), so:
\( r^4 = 2^4 \)
\( \implies r = 2 \)
Now, we can find the three geometric means:
\( G_1 = T_2 = ar = 16 \times 2 = 32 \)
\( G_2 = T_3 = ar^2 = 16 \times 2^2 = 16 \times 4 = 64 \)
\( G_3 = T_4 = ar^3 = 16 \times 2^3 = 16 \times 8 = 128 \)
Therefore, the three geometric means between 16 and 256 are 32, 64, and 128.
In simple words: We need to find three numbers that fit in a multiplication pattern between 16 and 256. We figured out that each number is twice the one before it. So, the three numbers are 32, 64, and 128.

🎯 Exam Tip: When inserting \( n \) geometric means between two numbers, the total number of terms in the G.P. becomes \( n+2 \).

Question 6. Insert 5 geometric means between \( \frac { 1 }{ 3 } \) and 243.
Answer:
To insert 5 geometric means (GM) between \( \frac{1}{3} \) and 243 means to form a Geometric Progression (G.P.) where \( \frac{1}{3} \) is the first term and 243 is the seventh term, with five terms in between them.
Let the G.P. be \( \frac{1}{3}, G_1, G_2, G_3, G_4, G_5, 243 \).
Here, the first term \( a = \frac{1}{3} \).
The seventh term \( T_7 = 243 \).
Using the G.P. formula \( T_n = ar^{n-1} \):
\( T_7 = ar^{7-1} = ar^6 \)
Substitute the known values:
\( 243 = \frac{1}{3} \times r^6 \)
Multiply by 3:
\( r^6 = 243 \times 3 \)
\( r^6 = 729 \)
We know that \( 3^6 = 729 \), so:
\( r^6 = 3^6 \)
\( \implies r = 3 \)
Now, we can find the five geometric means:
\( G_1 = T_2 = ar = \frac{1}{3} \times 3 = 1 \)
\( G_2 = T_3 = ar^2 = \frac{1}{3} \times 3^2 = \frac{1}{3} \times 9 = 3 \)
\( G_3 = T_4 = ar^3 = \frac{1}{3} \times 3^3 = \frac{1}{3} \times 27 = 9 \)
\( G_4 = T_5 = ar^4 = \frac{1}{3} \times 3^4 = \frac{1}{3} \times 81 = 27 \)
\( G_5 = T_6 = ar^5 = \frac{1}{3} \times 3^5 = \frac{1}{3} \times 243 = 81 \)
Therefore, the five geometric means between \( \frac{1}{3} \) and 243 are 1, 3, 9, 27, and 81.
In simple words: We are filling in five numbers to make a multiplying pattern between \( \frac{1}{3} \) and 243. We found the multiplication factor is 3. So, the numbers are 1, 3, 9, 27, and 81.

🎯 Exam Tip: Be careful with calculations involving fractions and powers to avoid errors, especially when the common ratio is a whole number or a fraction.

Question 7. If the A.M. and G.M. between two numbers are respectively 17 and 8, find the numbers.
Answer:
Let the two unknown numbers be \( a \) and \( b \).
The Arithmetic Mean (A.M.) of \( a \) and \( b \) is given as 17:
\( \frac{a+b}{2} = 17 \)
\( \implies a+b = 34 \) (Equation 1)
The Geometric Mean (G.M.) of \( a \) and \( b \) is given as 8:
\( \sqrt{ab} = 8 \)
Square both sides to eliminate the square root:
\( ab = 8^2 \)
\( \implies ab = 64 \) (Equation 2)
We use the algebraic identity \( (a-b)^2 = (a+b)^2 - 4ab \). This identity helps us find the difference between \( a \) and \( b \).
Substitute the values from Equation 1 and Equation 2 into the identity:
\( (a-b)^2 = (34)^2 - 4 \times 64 \)
\( (a-b)^2 = 1156 - 256 \)
\( (a-b)^2 = 900 \)
Take the square root of both sides:
\( a-b = \pm \sqrt{900} \)
\( \implies a-b = \pm 30 \)
This gives two possible cases:
**Case 1: When \( a-b = 30 \)** (Equation 3)
Add Equation 1 and Equation 3:
\( (a+b) + (a-b) = 34 + 30 \)
\( 2a = 64 \)
\( \implies a = \frac{64}{2} \)
\( \implies a = 32 \)
Substitute \( a=32 \) into Equation 1:
\( 32 + b = 34 \)
\( \implies b = 34 - 32 \)
\( \implies b = 2 \)
**Case 2: When \( a-b = -30 \)** (Equation 4)
Add Equation 1 and Equation 4:
\( (a+b) + (a-b) = 34 + (-30) \)
\( 2a = 4 \)
\( \implies a = \frac{4}{2} \)
\( \implies a = 2 \)
Substitute \( a=2 \) into Equation 1:
\( 2 + b = 34 \)
\( \implies b = 34 - 2 \)
\( \implies b = 32 \)
In both cases, the two numbers are 2 and 32.
In simple words: We were looking for two numbers. Their average is 17, and if you multiply them and then take the square root, you get 8. By using these facts, we found that the two numbers are 2 and 32.

🎯 Exam Tip: This problem often involves solving a system of equations, and the identity \( (a-b)^2 = (a+b)^2 - 4ab \) is a key shortcut to relate the sum and product of numbers to their difference.

Question 8. The second, third and sixth terms of an A.P. are consecutive terms of a geometric progression. Find the common ratio of the geometric progression.
Answer:
Let the first term of the Arithmetic Progression (A.P.) be \( a \) and the common difference be \( d \).
The general formula for the \( n^{th} \) term of an A.P. is \( T_n = a + (n-1)d \).
The second term of the A.P. is \( T_2 = a + (2-1)d = a+d \).
The third term of the A.P. is \( T_3 = a + (3-1)d = a+2d \).
The sixth term of the A.P. is \( T_6 = a + (6-1)d = a+5d \).
We are given that \( T_2, T_3, T_6 \) are consecutive terms of a Geometric Progression (G.P.).
For three consecutive terms \( X, Y, Z \) in a G.P., the property is \( Y^2 = XZ \).
So, \( (T_3)^2 = T_2 \times T_6 \)
\( (a+2d)^2 = (a+d)(a+5d) \)
Expand both sides:
\( a^2 + 4ad + 4d^2 = a^2 + 5ad + ad + 5d^2 \)
\( a^2 + 4ad + 4d^2 = a^2 + 6ad + 5d^2 \)
Subtract \( a^2 \) from both sides:
\( 4ad + 4d^2 = 6ad + 5d^2 \)
Move all terms to one side to solve for the relationship between \( a \) and \( d \):
\( 0 = 6ad - 4ad + 5d^2 - 4d^2 \)
\( 0 = 2ad + d^2 \)
Factor out \( d \):
\( 0 = d(2a + d) \)
This equation gives two possibilities:
1. \( d = 0 \): If the common difference is 0, all terms in the A.P. are the same (e.g., \( a, a, a, ... \)). Then \( T_2=a, T_3=a, T_6=a \), and the G.P. would be \( a, a, a \). The common ratio \( r = \frac{a}{a} = 1 \).
2. \( 2a + d = 0 \): This implies \( d = -2a \) (assuming \( d \neq 0 \)).
If \( d = -2a \), we can find the common ratio \( r \) of the G.P. using \( r = \frac{T_3}{T_2} \):
\( r = \frac{a+2d}{a+d} \)
Substitute \( d = -2a \) into this expression:
\( r = \frac{a+2(-2a)}{a+(-2a)} \)
\( r = \frac{a-4a}{a-2a} \)
\( r = \frac{-3a}{-a} \)
\( r = 3 \)
Thus, the common ratio of the geometric progression is 3 (assuming \( a \neq 0 \), otherwise \( d \) would also be 0).
In simple words: We used numbers from an adding pattern (A.P.) to form a multiplying pattern (G.P.). By using the rules for both types of patterns, we found a connection between the start of the A.P. and how much it changes. This connection showed us that the multiplication factor (common ratio) for the G.P. is 3.

🎯 Exam Tip: Clearly define your variables for the A.P. (first term 'a', common difference 'd') and then express the required terms using these variables before applying the G.P. property.

Question 9. If the \( l^{th} \), \( m^{th} \) and \( n^{th} \) terms of a G.P. are P, Q and S respectively. Show that \( Q^2 = PS \).
Answer:
Let \( A \) be the first term and \( R \) be the common ratio of the Geometric Progression (G.P.).
The general formula for the \( k^{th} \) term of a G.P. is \( T_k = AR^{k-1} \).
According to the problem:
The \( l^{th} \) term is \( P \): \( P = AR^{l-1} \) (Equation 1)
The \( m^{th} \) term is \( Q \): \( Q = AR^{m-1} \) (Equation 2)
The \( n^{th} \) term is \( S \): \( S = AR^{n-1} \) (Equation 3)
We need to show that \( Q^2 = PS \). This property holds true if the indices \( l, m, n \) are in an Arithmetic Progression, meaning \( m-l = n-m \), or \( 2m = l+n \). While not explicitly stated, the proof structure relies on this relationship.
Let's calculate the product \( PS \):
\( PS = (AR^{l-1}) \times (AR^{n-1}) \)
\( \implies PS = A \cdot A \cdot R^{(l-1)+(n-1)} \)
\( \implies PS = A^2 R^{l+n-2} \)
Now, let's calculate \( Q^2 \):
\( Q^2 = (AR^{m-1})^2 \)
\( \implies Q^2 = A^2 (R^{m-1})^2 \)
\( \implies Q^2 = A^2 R^{2(m-1)} \)
\( \implies Q^2 = A^2 R^{2m-2} \)
Since the problem implies that \( T_l, T_m, T_n \) have the G.P. property \( Q^2=PS \), it means that the exponents \( l, m, n \) are in A.P., i.e., \( l+n = 2m \).
Using this, we can see that \( l+n-2 = 2m-2 \).
Therefore, \( PS = A^2 R^{l+n-2} = A^2 R^{2m-2} = Q^2 \).
Hence, \( Q^2 = PS \) is proven.
In simple words: When you have three terms in a multiplying pattern (G.P.) that are at positions \( l, m, n \) in the sequence, and these positions themselves form an adding pattern, then the square of the middle term (\( Q \)) will always be equal to the first term (\( P \)) multiplied by the last term (\( S \)).

🎯 Exam Tip: This proof relies on the general properties of exponents in a G.P. and implicitly assumes that the positions \( l, m, n \) are themselves in an A.P. for the relationship to hold.

Question 10. The (p + q)th term and (p – q)th terms of a G.P. are a and b respectively. Find the pth term.
Answer:
Let \( A \) be the first term and \( R \) be the common ratio of the Geometric Progression (G.P.).
The general formula for the \( n^{th} \) term of a G.P. is \( T_n = AR^{n-1} \).
We are given the \( (p+q)^{th} \) term is \( a \):
\( T_{p+q} = AR^{(p+q)-1} = a \) (Equation 1)
We are given the \( (p-q)^{th} \) term is \( b \):
\( T_{p-q} = AR^{(p-q)-1} = b \) (Equation 2)
To find the common ratio \( R \), divide Equation 1 by Equation 2:
\( \frac{AR^{p+q-1}}{AR^{p-q-1}} = \frac{a}{b} \)
Using the exponent rule \( \frac{x^m}{x^n} = x^{m-n} \):
\( R^{(p+q-1) - (p-q-1)} = \frac{a}{b} \)
\( \implies R^{p+q-1-p+q+1} = \frac{a}{b} \)
\( \implies R^{2q} = \frac{a}{b} \)
Raise both sides to the power of \( \frac{1}{2q} \) to find \( R \):
\( R = \left(\frac{a}{b}\right)^{\frac{1}{2q}} \)
Now, we need to find the \( p^{th} \) term, \( T_p = AR^{p-1} \).
From Equation 1, we can express \( A \) as \( A = \frac{a}{R^{p+q-1}} \).
Substitute this expression for \( A \) into the formula for \( T_p \):
\( T_p = \frac{a}{R^{p+q-1}} \times R^{p-1} \)
Using the exponent rule \( \frac{x^m}{x^n} = x^{m-n} \):
\( T_p = a \times R^{(p-1) - (p+q-1)} \)
\( \implies T_p = a \times R^{p-1-p-q+1} \)
\( \implies T_p = a \times R^{-q} \)
Now, substitute the value of \( R \) we found:
\( T_p = a \times \left(\left(\frac{a}{b}\right)^{\frac{1}{2q}}\right)^{-q} \)
Using the exponent rule \( (x^m)^n = x^{mn} \):
\( T_p = a \times \left(\frac{a}{b}\right)^{\frac{1}{2q} \times (-q)} \)
\( \implies T_p = a \times \left(\frac{a}{b}\right)^{-\frac{q}{2q}} \)
\( \implies T_p = a \times \left(\frac{a}{b}\right)^{-\frac{1}{2}} \)
Using the exponent rule \( x^{-k} = \frac{1}{x^k} = \left(\frac{1}{x}\right)^k \):
\( T_p = a \times \left(\frac{b}{a}\right)^{\frac{1}{2}} \)
This can be written as:
\( T_p = a \times \frac{\sqrt{b}}{\sqrt{a}} \)
\( \implies T_p = \sqrt{a} \times \sqrt{a} \times \frac{\sqrt{b}}{\sqrt{a}} \)
\( \implies T_p = \sqrt{a}\sqrt{b} \)
\( \implies T_p = \sqrt{ab} \)
Therefore, the \( p^{th} \) term of the G.P. is \( \sqrt{ab} \).
In simple words: If the \( (p+q)^{th} \) term of a multiplying pattern is `a`, and the \( (p-q)^{th} \) term is `b`, then the \( p^{th} \) term of that pattern is found by multiplying `a` and `b` together and then taking the square root.

🎯 Exam Tip: When dealing with exponents, remember that \( (x^m)^n = x^{mn} \) and \( x^{-k} = 1/x^k \) are crucial for simplifying expressions and reaching the final solution.

Question 11. If the pth, qth, rth terms of a G.P. are x, y, z respectively, prove that xq-r, yr-p, zp-q = 1.
Answer:
Let \( A \) be the first term and \( R \) be the common ratio of the Geometric Progression (G.P.).
The general formula for the \( k^{th} \) term of a G.P. is \( T_k = AR^{k-1} \).
According to the problem:
The \( p^{th} \) term is \( x \): \( x = AR^{p-1} \) (Equation 1)
The \( q^{th} \) term is \( y \): \( y = AR^{q-1} \) (Equation 2)
The \( r^{th} \) term is \( z \): \( z = AR^{r-1} \) (Equation 3)
We need to prove that \( x^{q-r} y^{r-p} z^{p-q} = 1 \).
Let's consider the Left Hand Side (L.H.S.) of the equation:
L.H.S. \( = x^{q-r} y^{r-p} z^{p-q} \)
Substitute the expressions for \( x, y, z \) from Equations 1, 2, and 3 into the L.H.S.:
\( = (AR^{p-1})^{q-r} (AR^{q-1})^{r-p} (AR^{r-1})^{p-q} \)
Apply the exponent rule \( (uv)^k = u^k v^k \):
\( = A^{q-r} (R^{p-1})^{q-r} \times A^{r-p} (R^{q-1})^{r-p} \times A^{p-q} (R^{r-1})^{p-q} \)
Apply the exponent rule \( (u^m)^n = u^{mn} \):
\( = A^{q-r} R^{(p-1)(q-r)} \times A^{r-p} R^{(q-1)(r-p)} \times A^{p-q} R^{(r-1)(p-q)} \)
Group the terms with \( A \) and terms with \( R \) together:
\( = A^{(q-r)+(r-p)+(p-q)} \times R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)} \)
Now, simplify the exponent of \( A \):
\( (q-r) + (r-p) + (p-q) = q - r + r - p + p - q = 0 \)
So, the \( A \) part becomes \( A^0 \).
Next, simplify the exponent of \( R \):
Let's expand each part:
\( (p-1)(q-r) = pq - pr - q + r \)
\( (q-1)(r-p) = qr - qp - r + p \)
\( (r-1)(p-q) = rp - rq - p + q \)
Add these three expanded terms:
\( (pq - pr - q + r) + (qr - pq - r + p) + (pr - qr - p + q) \)
Observe that all terms cancel each other out (e.g., \( pq \) cancels with \( -pq \), \( -pr \) cancels with \( pr \), etc.).
The sum of these exponents is 0.
So, the \( R \) part becomes \( R^0 \).
Therefore, L.H.S. \( = A^0 R^0 = 1 \times 1 = 1 \).
This is equal to the Right Hand Side (R.H.S.).
Hence, \( x^{q-r} y^{r-p} z^{p-q} = 1 \) is proven.
In simple words: This problem shows a special math trick. If you have three numbers in a multiplying pattern at certain positions, and you raise each number to a power based on the difference between the other positions, then when you multiply these results, you will always get 1.

🎯 Exam Tip: When proving identities involving exponents, carefully expand and combine terms, ensuring that the exponents of the base terms simplify to zero for the desired result of 1.

Question 12. In a set of four numbers, the first three are in G.P. and the last three are in A.P. with difference 6. If the first number is the same as the fourth, find the four numbers.
Answer:
Let the four numbers be \( N_1, N_2, N_3, N_4 \).
We are given the following conditions:
1. The first three numbers (\( N_1, N_2, N_3 \)) are in a Geometric Progression (G.P.).
2. The last three numbers (\( N_2, N_3, N_4 \)) are in an Arithmetic Progression (A.P.) with a common difference of 6.
3. The first number is the same as the fourth (\( N_1 = N_4 \)).
From condition 2, let \( N_3 = a \). Since the common difference is 6, the terms are:
\( N_2 = a - 6 \)
\( N_3 = a \)
\( N_4 = a + 6 \)
From condition 3, \( N_1 = N_4 \), so \( N_1 = a+6 \).
Now, we have the four numbers expressed in terms of \( a \):
\( N_1 = a+6 \)
\( N_2 = a-6 \)
\( N_3 = a \)
\( N_4 = a+6 \)
From condition 1, the first three numbers \( (a+6), (a-6), a \) are in G.P.
For three terms \( X, Y, Z \) to be in G.P., the property is \( Y^2 = XZ \).
So, \( (a-6)^2 = (a+6) \times a \)
Expand both sides:
\( a^2 - 12a + 36 = a^2 + 6a \)
Subtract \( a^2 \) from both sides:
\( -12a + 36 = 6a \)
Add \( 12a \) to both sides:
\( 36 = 6a + 12a \)
\( 36 = 18a \)
Solve for \( a \):
\( a = \frac{36}{18} \)
\( \implies a = 2 \)
Now, substitute \( a=2 \) back into the expressions for the four numbers:
\( N_1 = a+6 = 2+6 = 8 \)
\( N_2 = a-6 = 2-6 = -4 \)
\( N_3 = a = 2 \)
\( N_4 = a+6 = 2+6 = 8 \)
So, the four numbers are 8, -4, 2, 8.
Let's verify the conditions:
* First three in G.P.: \( 8, -4, 2 \). The common ratio is \( \frac{-4}{8} = -\frac{1}{2} \) and \( \frac{2}{-4} = -\frac{1}{2} \). (Correct)
* Last three in A.P. with difference 6: \( -4, 2, 8 \). The common difference is \( 2 - (-4) = 6 \) and \( 8 - 2 = 6 \). (Correct)
* First number equals fourth: \( 8 = 8 \). (Correct)
In simple words: We found a list of four numbers that fit all the rules given. The numbers are 8, -4, 2, and 8. The first three numbers follow a multiplying pattern, and the last three follow an adding pattern with a difference of 6, and the first number is the same as the last.

🎯 Exam Tip: When problems involve both A.P. and G.P. properties, define the terms systematically using one variable and then apply the respective definitions to form an equation.

Question 13. If \( a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}} \) and a, b, c, are in G.P., prove that x, y, z are in A.P.
Answer:
Given that \( a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} \).
Let this common value be \( k \), where \( k \neq 0 \).
So, we have:
\( a^{\frac{1}{x}} = k \implies a = k^x \)
\( b^{\frac{1}{y}} = k \implies b = k^y \)
\( c^{\frac{1}{z}} = k \implies c = k^z \)
We are also given that \( a, b, c \) are in a Geometric Progression (G.P.).
For three terms \( a, b, c \) to be in G.P., the property is that the square of the middle term is equal to the product of the first and third terms:
\( b^2 = ac \)
Now, substitute the expressions for \( a, b, c \) in terms of \( k \):
\( (k^y)^2 = (k^x) \times (k^z) \)
Using the exponent rules \( (u^m)^n = u^{mn} \) and \( u^m \times u^n = u^{m+n} \):
\( k^{2y} = k^{x+z} \)
Since the bases are equal and positive (as \( k \neq 0 \)), their exponents must be equal:
\( 2y = x+z \)
This equation, \( 2y = x+z \), is the defining condition for three numbers \( x, y, z \) to be in an Arithmetic Progression (A.P.).
Therefore, \( x, y, z \) are in A.P.
This completes the proof.
In simple words: If three numbers \( a, b, c \) follow a multiplying pattern, and they are related to other numbers \( x, y, z \) through powers, then those numbers \( x, y, z \) will follow an adding pattern.

🎯 Exam Tip: When given equations with multiple equalities, introducing a common constant \( k \) simplifies the relationships between variables, making it easier to apply progression properties.

Question 14. If one G.M., G and two A.M's p and q be inserted between two given numbers, prove that G2 = (2p - q)(2q – p).
Answer:
Let the two given numbers be \( a \) and \( b \).
We are given that \( G \) is the Geometric Mean (G.M.) between \( a \) and \( b \).
By definition, \( G = \sqrt{ab} \). Squaring both sides gives:
\( G^2 = ab \) (Equation 1)
We are also given that \( p \) and \( q \) are two Arithmetic Means (A.M.s) inserted between \( a \) and \( b \).
This means that \( a, p, q, b \) form an Arithmetic Progression (A.P.).
Let \( d \) be the common difference of this A.P.
In this A.P., \( a \) is the first term \( (T_1) \) and \( b \) is the fourth term \( (T_4) \).
Using the A.P. formula \( T_n = T_1 + (n-1)d \):
\( T_4 = a + (4-1)d \)
\( b = a+3d \)
From this, we can find the common difference \( d \):
\( 3d = b-a \)
\( \implies d = \frac{b-a}{3} \)
Now, we can express \( p \) and \( q \) in terms of \( a \) and \( d \):
\( p = T_2 = a+d = a + \frac{b-a}{3} \)
\( \implies p = \frac{3a + b-a}{3} = \frac{2a+b}{3} \) (Equation 2)
\( q = T_3 = a+2d = a + 2\left(\frac{b-a}{3}\right) \)
\( \implies q = \frac{3a + 2b-2a}{3} = \frac{a+2b}{3} \) (Equation 3)
We need to prove that \( G^2 = (2p - q)(2q - p) \). Let's evaluate the Right Hand Side (R.H.S.) using the expressions for \( p \) and \( q \):
R.H.S. \( = (2p - q)(2q - p) \)
Substitute \( p = \frac{2a+b}{3} \) and \( q = \frac{a+2b}{3} \):
\( = \left(2\left(\frac{2a+b}{3}\right) - \left(\frac{a+2b}{3}\right)\right) \left(2\left(\frac{a+2b}{3}\right) - \left(\frac{2a+b}{3}\right)\right) \)
Combine the terms within each parenthesis:
\( = \left(\frac{2(2a+b) - (a+2b)}{3}\right) \left(\frac{2(a+2b) - (2a+b)}{3}\right) \)
\( = \left(\frac{4a+2b-a-2b}{3}\right) \left(\frac{2a+4b-2a-b}{3}\right) \)
Simplify the numerators:
\( = \left(\frac{3a}{3}\right) \left(\frac{3b}{3}\right) \)
\( = (a)(b) \)
\( = ab \)
From Equation 1, we know that \( G^2 = ab \).
Therefore, L.H.S. \( = G^2 \) and R.H.S. \( = ab \). Since \( G^2 = ab \), we have proven that \( G^2 = (2p - q)(2q - p) \).
In simple words: When a geometric mean (G) and two arithmetic means (p and q) are placed between two numbers, it means that the square of G is equal to the result of a special multiplication: `(twice p minus q)` multiplied by `(twice q minus p)`.

🎯 Exam Tip: Systematically express all inserted means in terms of the original numbers and the common difference/ratio. This helps in substituting them correctly into the equation to be proved.

Question 15. Construct a quadratic equation in x such that the A.M. of its roots is A and G.M. is G.
Answer:
Let the roots of the required quadratic equation be \( \alpha \) and \( \beta \).
We are given that the Arithmetic Mean (A.M.) of its roots is \( A \).
\( \frac{\alpha+\beta}{2} = A \)
Multiplying by 2, we get the sum of the roots:
\( \alpha+\beta = 2A \)
We are also given that the Geometric Mean (G.M.) of its roots is \( G \).
\( \sqrt{\alpha\beta} = G \)
Squaring both sides, we get the product of the roots:
\( \alpha\beta = G^2 \)
The general form of a quadratic equation with roots \( \alpha \) and \( \beta \) is given by:
\( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \)
Substitute the values we found for the sum of roots and product of roots:
\( x^2 - (2A)x + (G^2) = 0 \)
\( \implies x^2 - 2Ax + G^2 = 0 \)
This is the required quadratic equation.
In simple words: If you know the average of the answers (roots) to a quadratic equation, and also their geometric mean (multiplied then square-rooted), you can quickly write down the equation itself. It will look like \( x^2 \) minus `(twice the average)` times `x`, plus `(the geometric mean squared)`, all equal to zero.

🎯 Exam Tip: Remember the fundamental relationship between the roots of a quadratic equation and its coefficients: \( x^2 - (\alpha+\beta)x + \alpha\beta = 0 \).

Question 16. The fourth term of a G.P. is greater than the first term, which is positive, by 372 . The third term is greater than the second by 60 . Calculate the common ratio and the first term of the progression.
Answer:
Let the first term of the Geometric Progression (G.P.) be \( a \) and the common ratio be \( r \). We are given that \( a \) is positive.
The general formula for the \( n^{th} \) term of a G.P. is \( T_n = ar^{n-1} \).
From the first condition: "The fourth term of a G.P. is greater than the first term by 372."
\( T_4 = T_1 + 372 \)
\( ar^3 = a + 372 \)
Rearrange the terms:
\( ar^3 - a = 372 \)
Factor out \( a \):
\( a(r^3 - 1) = 372 \) (Equation 1)
From the second condition: "The third term is greater than the second by 60."
\( T_3 = T_2 + 60 \)
\( ar^2 = ar + 60 \)
Rearrange the terms:
\( ar^2 - ar = 60 \)
Factor out \( ar \):
\( ar(r - 1) = 60 \) (Equation 2)
Now, divide Equation 1 by Equation 2 to eliminate \( a \) and find \( r \):
\( \frac{a(r^3 - 1)}{ar(r - 1)} = \frac{372}{60} \)
Cancel \( a \) and use the algebraic identity \( r^3 - 1 = (r-1)(r^2+r+1) \):
\( \frac{(r-1)(r^2+r+1)}{r(r-1)} = \frac{31 \times 12}{5 \times 12} \)
Assuming \( r \neq 1 \), we can cancel \( (r-1) \):
\( \implies \frac{r^2+r+1}{r} = \frac{31}{5} \)
Cross-multiply:
\( 5(r^2+r+1) = 31r \)
\( 5r^2 + 5r + 5 = 31r \)
Rearrange into a quadratic equation:
\( 5r^2 + 5r - 31r + 5 = 0 \)
\( 5r^2 - 26r + 5 = 0 \)
Solve this quadratic equation by factoring:
\( 5r^2 - 25r - r + 5 = 0 \)
\( 5r(r-5) - 1(r-5) = 0 \)
\( (5r-1)(r-5) = 0 \)
This gives two possible values for \( r \):
1. \( 5r-1 = 0 \implies r = \frac{1}{5} \)
2. \( r-5 = 0 \implies r = 5 \)
Now, we find the first term \( a \) for each value of \( r \) using Equation 1:
**Case 1: If \( r = 5 \)**
Substitute \( r=5 \) into Equation 1:
\( a(5^3 - 1) = 372 \)
\( a(125 - 1) = 372 \)
\( a(124) = 372 \)
\( a = \frac{372}{124} \)
\( \implies a = 3 \)
This value of \( a=3 \) is positive, which satisfies the problem's condition.
**Case 2: If \( r = \frac{1}{5} \)**
Substitute \( r=\frac{1}{5} \) into Equation 1:
\( a\left(\left(\frac{1}{5}\right)^3 - 1\right) = 372 \)
\( a\left(\frac{1}{125} - 1\right) = 372 \)
\( a\left(\frac{1-125}{125}\right) = 372 \)
\( a\left(\frac{-124}{125}\right) = 372 \)
\( a = 372 \times \frac{125}{-124} \)
\( a = -3 \times 125 \)
\( \implies a = -375 \)
This value of \( a=-375 \) is negative, which contradicts the problem's condition that the first term is positive. Therefore, this case is not valid.
Thus, the common ratio is 5 and the first term is 3.
In simple words: We used the given clues about a multiplying pattern to find two possible ways the pattern could work. One way made the first number negative, which was not allowed. The other way showed the first number is 3 and each number is multiplied by 5 to get the next one.

🎯 Exam Tip: Always double-check solutions against all given conditions, such as the positivity of a term, to filter out extraneous results from algebraic manipulations.

Question 17. The first, eighth and twenty-second terms of an A.P. are three consecutive terms of a G.P. Find the common ratio of the G.P. Given also that the sum of the first twenty-two terms of the A.P. is 275, find its first term.
Answer:
Let the first term of the Arithmetic Progression (A.P.) be \( a \) and the common difference be \( d \).
The general formula for the \( n^{th} \) term of an A.P. is \( T_n = a + (n-1)d \).
The terms of the A.P. are:
\( T_1 = a \)
\( T_8 = a + (8-1)d = a + 7d \)
\( T_{22} = a + (22-1)d = a + 21d \)
We are given that \( T_1, T_8, T_{22} \) are three consecutive terms of a Geometric Progression (G.P.).
For three terms \( X, Y, Z \) to be in G.P., the property is \( Y^2 = XZ \).
So, \( (T_8)^2 = T_1 \times T_{22} \)
\( (a + 7d)^2 = a(a + 21d) \)
Expand both sides:
\( a^2 + 14ad + 49d^2 = a^2 + 21ad \)
Subtract \( a^2 \) from both sides:
\( 14ad + 49d^2 = 21ad \)
Rearrange the terms to find a relationship between \( a \) and \( d \):
\( 49d^2 = 21ad - 14ad \)
\( 49d^2 = 7ad \)
Move all terms to one side:
\( 49d^2 - 7ad = 0 \)
Factor out \( 7d \):
\( 7d(7d - a) = 0 \)
This equation gives two possibilities:
1. \( d = 0 \): If the common difference is 0, all terms in the A.P. are the same. Then \( T_1=a, T_8=a, T_{22}=a \), and the G.P. common ratio would be 1.
2. \( 7d - a = 0 \): This implies \( a = 7d \) (assuming \( d \neq 0 \)). (Equation 1)

**Part 1: Find the common ratio of the G.P.**
The common ratio \( r \) of the G.P. is \( \frac{T_8}{T_1} \).
\( r = \frac{a+7d}{a} \)
Substitute \( a = 7d \) from Equation 1 into this expression:
\( r = \frac{7d+7d}{7d} = \frac{14d}{7d} \)
\( \implies r = 2 \)
The common ratio of the G.P. is 2.

**Part 2: Find the first term of the A.P.**
We are given that the sum of the first twenty-two terms of the A.P. is 275.
The formula for the sum of \( n \) terms of an A.P. is \( S_n = \frac{n}{2}(2a + (n-1)d) \).
For \( n=22 \):
\( S_{22} = \frac{22}{2}(2a + (22-1)d) \)
\( S_{22} = 11(2a + 21d) \)
We are given \( S_{22} = 275 \):
\( 11(2a + 21d) = 275 \)
Divide both sides by 11:
\( 2a + 21d = \frac{275}{11} \)
\( \implies 2a + 21d = 25 \) (Equation 2)
Now, substitute \( a = 7d \) (from Equation 1) into Equation 2:
\( 2(7d) + 21d = 25 \)
\( 14d + 21d = 25 \)
\( 35d = 25 \)
\( \implies d = \frac{25}{35} = \frac{5}{7} \)
Finally, find \( a \) using \( a = 7d \):
\( a = 7 \times \frac{5}{7} \)
\( \implies a = 5 \)
The first term of the A.P. is 5, and the common ratio of the G.P. is 2.
In simple words: We had an adding pattern (A.P.) where its 1st, 8th, and 22nd numbers also formed a multiplying pattern (G.P.). We used this to find the G.P.'s multiplication factor, which is 2. We also knew the sum of the first 22 numbers in the A.P. is 275, which helped us find that the very first number in the A.P. is 5.

🎯 Exam Tip: Pay close attention to distinguishing between terms of an A.P. and terms of a G.P., especially when they are derived from each other, to apply the correct formulas and properties.