OP Malhotra Class 10 Maths Solutions Chapter 9 Arithmetic and Geometric Progression Exercise 9 (C)

S Chand Class 10 ICSE Maths Solutions Chapter 9 Arithmetic And Geometric Progression Ex 9(C)

Question 1. Determine whether the following sequences are geometric progressions or not? If yes then find the common ratio.
(i) 27, 9, 3, 1, ...
(ii) -1, 2, 4, 8
(iii) 2, \( \frac{1}{2}, \frac{1}{8}, \frac{1}{32}, \) ...
(iv) -12, -6, 0, 6, ...
Answer:
(i) Given sequence: 27, 9, 3, 1, ...
The first term is \( a = 27 \).
To find the common ratio \( r \), we divide the second term by the first: \( r = \frac{9}{27} = \frac{1}{3} \).
Next, we check the ratio of the third term to the second: \( \frac{3}{9} = \frac{1}{3} \).
Then, the ratio of the fourth term to the third: \( \frac{1}{3} \).
Since the ratio between consecutive terms is constant, this sequence is a Geometric Progression (G.P.). The common ratio \( r \) is \( \frac{1}{3} \). A geometric progression has a constant ratio between consecutive terms, which is crucial for predicting future terms.
(ii) Given sequence: -1, 2, 4, 8
The first term is \( a = -1 \).
The ratio of the second term to the first is \( r = \frac{2}{-1} = -2 \).
The ratio of the third term to the second is \( \frac{4}{2} = 2 \).
Since the ratios \( -2 \) and \( 2 \) are not the same, the common ratio is not constant. Therefore, this sequence is not a Geometric Progression. For a sequence to be a G.P., every pair of consecutive terms must have the same ratio.
(iii) Given sequence: 2, \( \frac{1}{2}, \frac{1}{8}, \frac{1}{32}, \) ...
The first term is \( a = 2 \).
The ratio of the second term to the first is \( r = \frac{1}{2} \div 2 = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).
The ratio of the third term to the second is \( \frac{1}{8} \div \frac{1}{2} = \frac{1}{8} \times \frac{2}{1} = \frac{1}{4} \).
The ratio of the fourth term to the third is \( \frac{1}{32} \div \frac{1}{8} = \frac{1}{32} \times \frac{8}{1} = \frac{1}{4} \).
Since the ratio is constant, this sequence is a Geometric Progression (G.P.). The common ratio \( r \) is \( \frac{1}{4} \). Dividing by a number is the same as multiplying by its reciprocal.
(iv) Given sequence: -12, -6, 0, 6, ...
The first term is \( a = -12 \).
The ratio of the second term to the first is \( r = \frac{-6}{-12} = \frac{1}{2} \).
The ratio of the third term to the second is \( \frac{0}{-6} = 0 \).
Since the ratios \( \frac{1}{2} \) and \( 0 \) are not the same, the common ratio is not constant. Therefore, this sequence is not a Geometric Progression. A sequence with a zero term cannot maintain a constant non-zero common ratio.
In simple words: To check if a sequence is a geometric progression, always divide each term by the one before it. If all these divisions give the same number, then it is a G.P., and that number is the common ratio. If the numbers you get are different, it's not a G.P.

🎯 Exam Tip: When determining if a sequence is a G.P., always calculate the ratio for at least two pairs of consecutive terms to confirm consistency. One ratio is not enough to prove it.

Question 2. Write the next three terms in each of the GPs given below:
(i) 2, 6, ...
(ii) \( \frac{1}{16}, -\frac{1}{8}, \) ...
(iii) 0.3, 0.06, ...
Answer:
(i) Given G.P.: 2, 6, ...
The first term is \( a = 2 \).
The common ratio is \( r = \frac{6}{2} = 3 \).
To find the next three terms, multiply the last given term by the common ratio:
Next term 1: \( 6 \times 3 = 18 \)
Next term 2: \( 18 \times 3 = 54 \)
Next term 3: \( 54 \times 3 = 162 \)
The next three terms are 18, 54, and 162. Finding the common ratio is the first crucial step to extending any geometric sequence.
(ii) Given G.P.: \( \frac{1}{16}, -\frac{1}{8}, \) ...
The first term is \( a = \frac{1}{16} \).
The common ratio is \( r = \left(-\frac{1}{8}\right) \div \left(\frac{1}{16}\right) = -\frac{1}{8} \times \frac{16}{1} = -2 \).
To find the next three terms:
Next term 1: \( -\frac{1}{8} \times (-2) = \frac{2}{8} = \frac{1}{4} \)
Next term 2: \( \frac{1}{4} \times (-2) = -\frac{2}{4} = -\frac{1}{2} \)
Next term 3: \( -\frac{1}{2} \times (-2) = 1 \)
The next three terms are \( \frac{1}{4}, -\frac{1}{2}, \) and \( 1 \). A negative common ratio causes the signs of the terms to alternate.
(iii) Given G.P.: 0.3, 0.06, ...
The first term is \( a = 0.3 \).
The common ratio is \( r = \frac{0.06}{0.3} = 0.2 \).
To find the next three terms:
Next term 1: \( 0.06 \times 0.2 = 0.012 \)
Next term 2: \( 0.012 \times 0.2 = 0.0024 \)
Next term 3: \( 0.0024 \times 0.2 = 0.00048 \)
The next three terms are 0.012, 0.0024, and 0.00048. Multiplying by a decimal less than 1 makes the terms progressively smaller.
In simple words: First, find the common ratio by dividing the second term by the first. Then, keep multiplying the last term you have by this ratio to get the next terms in the sequence.

🎯 Exam Tip: Always calculate the common ratio carefully. A small error in the ratio will lead to incorrect subsequent terms.

Question 3. Find the:
(i) 6th term of the G.P. 2, 10, 50 ...
(ii) 11th term of the GP. 4, 12, 36 ...
Answer:
(i) For the G.P. 2, 10, 50, ...
The first term is \( a = 2 \).
The common ratio is \( r = \frac{10}{2} = 5 \).
To find the 6th term (\( T_6 \)), we use the formula \( T_n = ar^{n-1} \).
\( T_6 = 2 \times 5^{6-1} \)
\( T_6 = 2 \times 5^5 \)
\( T_6 = 2 \times 3125 \)
\( T_6 = 6250 \)
The 6th term of this G.P. is 6250. The formula \( T_n = ar^{n-1} \) is a powerful tool to find any term in a geometric sequence without listing all previous terms.
(ii) For the G.P. 4, 12, 36, ...
The first term is \( a = 4 \).
The common ratio is \( r = \frac{12}{4} = 3 \).
To find the 11th term (\( T_{11} \)), we use the formula \( T_n = ar^{n-1} \).
\( T_{11} = 4 \times 3^{11-1} \)
\( T_{11} = 4 \times 3^{10} \)
\( T_{11} = 4 \times 59049 \)
\( T_{11} = 236196 \)
The 11th term of this G.P. is 236196. The terms grow very quickly in a geometric progression when the common ratio is greater than 1.
In simple words: Use the formula \( T_n = ar^{n-1} \), where \( a \) is the first number, \( r \) is the common ratio, and \( n \) is the term number you want to find. Just plug in the values and calculate.

🎯 Exam Tip: Remember to calculate the common ratio (r) correctly first. Be careful with exponents, especially for larger 'n' values.

Question 4. Write the first five terms of the G.P. where nth term is given as:
(i) \( 4 \cdot 3^{n-1} \)
(ii) \( \frac{5^{n-1}}{2^{n+1}} \)
Answer:
(i) The formula for the \( n \)th term is \( T_n = 4 \cdot 3^{n-1} \). We need to find the first five terms.
For \( n = 1 \): \( T_1 = 4 \cdot 3^{1-1} = 4 \cdot 3^0 = 4 \cdot 1 = 4 \).
For \( n = 2 \): \( T_2 = 4 \cdot 3^{2-1} = 4 \cdot 3^1 = 4 \cdot 3 = 12 \).
For \( n = 3 \): \( T_3 = 4 \cdot 3^{3-1} = 4 \cdot 3^2 = 4 \cdot 9 = 36 \).
For \( n = 4 \): \( T_4 = 4 \cdot 3^{4-1} = 4 \cdot 3^3 = 4 \cdot 27 = 108 \).
For \( n = 5 \): \( T_5 = 4 \cdot 3^{5-1} = 4 \cdot 3^4 = 4 \cdot 81 = 324 \).
Thus, the first five terms of the G.P. are 4, 12, 36, 108, and 324. In this general term formula, 4 represents the first term and 3 is the common ratio.
(ii) The formula for the \( n \)th term is \( T_n = \frac{5^{n-1}}{2^{n+1}} \). We need to find the first five terms.
For \( n = 1 \): \( T_1 = \frac{5^{1-1}}{2^{1+1}} = \frac{5^0}{2^2} = \frac{1}{4} \).
For \( n = 2 \): \( T_2 = \frac{5^{2-1}}{2^{2+1}} = \frac{5^1}{2^3} = \frac{5}{8} \).
For \( n = 3 \): \( T_3 = \frac{5^{3-1}}{2^{3+1}} = \frac{5^2}{2^4} = \frac{25}{16} \).
For \( n = 4 \): \( T_4 = \frac{5^{4-1}}{2^{4+1}} = \frac{5^3}{2^5} = \frac{125}{32} \).
For \( n = 5 \): \( T_5 = \frac{5^{5-1}}{2^{5+1}} = \frac{5^4}{2^6} = \frac{625}{64} \).
Thus, the first five terms of the G.P. are \( \frac{1}{4}, \frac{5}{8}, \frac{25}{16}, \frac{125}{32}, \) and \( \frac{625}{64} \). Notice how both the numerator (powers of 5) and the denominator (powers of 2) grow with each term.
In simple words: To find the terms, just put the numbers 1, 2, 3, 4, and 5 into the formula for 'n'. Calculate each result to get the first five terms.

🎯 Exam Tip: When evaluating terms with exponents, ensure you follow the order of operations correctly, especially with fractions and negative signs.

Question 5. Write down the nth term of each of the following GPs whose first two terms are given as follows. Also find the term stated besides each G.P.
(i) 12, -36, ... sixth term
(ii) \( 3, - \frac{1}{3}, \) ... 8th term
(iii) \( b^2c^3, b^3c^2, \) ... 5th term
Answer:
(i) Given G.P.: 12, -36, ... and we need the 6th term.
The first term is \( a = 12 \).
The common ratio is \( r = \frac{-36}{12} = -3 \).
The formula for the \( n \)th term is \( T_n = ar^{n-1} \).
For the 6th term (\( T_6 \)):
\( T_6 = 12 \times (-3)^{6-1} \)
\( T_6 = 12 \times (-3)^5 \)
\( T_6 = 12 \times (-243) \)
\( T_6 = -2916 \)
The 6th term is -2916. A negative common ratio makes the terms alternate between positive and negative values.
(ii) Given G.P.: \( 3, - \frac{1}{3}, \) ... and we need the 8th term.
The first term is \( a = 3 \).
The common ratio is \( r = \left(-\frac{1}{3}\right) \div 3 = -\frac{1}{3} \times \frac{1}{3} = -\frac{1}{9} \).
The formula for the \( n \)th term is \( T_n = ar^{n-1} \).
For the 8th term (\( T_8 \)):
\( T_8 = 3 \times \left(-\frac{1}{9}\right)^{8-1} \)
\( T_8 = 3 \times \left(-\frac{1}{9}\right)^7 \)
\( T_8 = 3 \times \left(-\frac{1}{4782969}\right) \)
\( T_8 = -\frac{3}{4782969} = -\frac{1}{1594323} \)
The 8th term is \( -\frac{1}{1594323} \). When the absolute value of the common ratio is less than 1, the terms get closer to zero as \( n \) increases.
(iii) Given G.P.: \( b^2c^3, b^3c^2, \) ... and we need the 5th term.
The first term is \( a = b^2c^3 \).
The common ratio is \( r = \frac{b^3c^2}{b^2c^3} = \frac{b}{c} \).
The formula for the \( n \)th term is \( T_n = ar^{n-1} \).
For the 5th term (\( T_5 \)):
\( T_5 = b^2c^3 \times \left(\frac{b}{c}\right)^{5-1} \)
\( T_5 = b^2c^3 \times \left(\frac{b}{c}\right)^4 \)
\( T_5 = b^2c^3 \times \frac{b^4}{c^4} \)
\( T_5 = b^{2+4}c^{3-4} \)
\( T_5 = b^6c^{-1} \)
\( T_5 = \frac{b^6}{c} \)
The 5th term is \( \frac{b^6}{c} \). Algebraic expressions can also form geometric progressions, following the same rules for terms and ratios.
In simple words: First, find the starting term and the common ratio from the given terms. Then, use the general formula \( T_n = ar^{n-1} \) to find the specific term asked for.

🎯 Exam Tip: Pay close attention to negative signs and fractional exponents when calculating the terms, as these can easily lead to errors.

Question 6. Which term of the G.P. 27, -18,12, -8, ... is \( \frac{1024}{2187} \).
Answer:
The given Geometric Progression (G.P.) is 27, -18, 12, -8, ...
The first term is \( a = 27 \).
The common ratio is \( r = \frac{-18}{27} = -\frac{2}{3} \).
We want to find which term is equal to \( \frac{1024}{2187} \). Let this be the \( n \)th term, \( T_n \).
The formula for the \( n \)th term is \( T_n = ar^{n-1} \).
So, \( \frac{1024}{2187} = 27 \times \left(-\frac{2}{3}\right)^{n-1} \)
Now, divide both sides by 27:
\( \implies \left(-\frac{2}{3}\right)^{n-1} = \frac{1024}{2187 \times 27} \)
We need to express the right side as a power of \( -\frac{2}{3} \).
We know that \( 1024 = 2^{10} \), \( 2187 = 3^7 \), and \( 27 = 3^3 \).
So, \( \left(-\frac{2}{3}\right)^{n-1} = \frac{2^{10}}{3^7 \times 3^3} = \frac{2^{10}}{3^{10}} \)
\( \implies \left(-\frac{2}{3}\right)^{n-1} = \left(\frac{2}{3}\right)^{10} \)
Since \( 10 \) is an even exponent, \( \left(\frac{2}{3}\right)^{10} \) is the same as \( \left(-\frac{2}{3}\right)^{10} \).
\( \implies \left(-\frac{2}{3}\right)^{n-1} = \left(-\frac{2}{3}\right)^{10} \)
By comparing the exponents, we get:
\( \implies n - 1 = 10 \)
\( \implies n = 10 + 1 \)
\( \implies n = 11 \)
Therefore, \( \frac{1024}{2187} \) is the 11th term of the G.P. Identifying the correct powers is essential when comparing terms in an equation like this.
In simple words: First, find the first number and the common ratio of the G.P. Then, use the formula for the nth term, \( T_n = ar^{n-1} \), and solve for \( n \) by making sure both sides of the equation have the same base with exponents.

🎯 Exam Tip: When the common ratio is negative, be careful with the exponent. If the desired term is positive, the exponent \( (n-1) \) must be an even number.

Question 7. Write the GP. whose 4th term is 54 and the 7th term is 1458.
Answer:
In a Geometric Progression (G.P.), we are given:
The 4th term \( T_4 = 54 \).
The 7th term \( T_7 = 1458 \).
Let \( a \) be the first term and \( r \) be the common ratio. The formula for the \( n \)th term is \( T_n = ar^{n-1} \).
Using this, we can write:
\( T_4 = ar^{4-1} = ar^3 = 54 \) (Equation i)
\( T_7 = ar^{7-1} = ar^6 = 1458 \) (Equation ii)
To find \( r \), we divide Equation (ii) by Equation (i):
\( \implies \frac{ar^6}{ar^3} = \frac{1458}{54} \)
\( \implies r^{6-3} = 27 \)
\( \implies r^3 = 27 \)
Since \( 27 = 3^3 \), we have:
\( \implies r^3 = 3^3 \)
\( \implies r = 3 \)
Now that we have \( r = 3 \), we can substitute it back into Equation (i) to find \( a \):
\( ar^3 = 54 \)
\( a(3)^3 = 54 \)
\( a \times 27 = 54 \)
\( \implies a = \frac{54}{27} \)
\( \implies a = 2 \)
So, the first term is \( a = 2 \) and the common ratio is \( r = 3 \).
The Geometric Progression will be:
First term: \( 2 \)
Second term: \( 2 \times 3 = 6 \)
Third term: \( 6 \times 3 = 18 \)
Fourth term: \( 18 \times 3 = 54 \)
The G.P. is 2, 6, 18, 54, ... By using the ratio of two terms, we can efficiently find the common ratio and then the first term.
In simple words: First, set up two equations using the general term formula \( T_n = ar^{n-1} \) for the given terms. Divide the equations to find the common ratio \( r \). Then, use \( r \) in one of the equations to find the first term \( a \). Finally, write out the G.P. using \( a \) and \( r \).

🎯 Exam Tip: This type of problem requires solving a system of equations involving exponents. Remember to divide the higher-indexed term equation by the lower-indexed one to easily find 'r'.

Question 8. The last term of the G.P. : \( 3, 3\sqrt{3}, 9,... \) is 2187. How many terms in all there in the GP.?
Answer:
The given Geometric Progression (G.P.) is \( 3, 3\sqrt{3}, 9, ... \) and its last term is 2187.
The first term is \( a = 3 \).
The common ratio is \( r = \frac{3\sqrt{3}}{3} = \sqrt{3} \).
Let the last term, \( T_n \), be 2187. The formula for the \( n \)th term is \( T_n = ar^{n-1} \).
Substitute the known values:
\( 2187 = 3 \times (\sqrt{3})^{n-1} \)
Divide by 3:
\( \implies (\sqrt{3})^{n-1} = \frac{2187}{3} \)
\( \implies (\sqrt{3})^{n-1} = 729 \)
We need to express 729 as a power of \( \sqrt{3} \).
We know that \( 729 = 3^6 \).
Since \( 3 = (\sqrt{3})^2 \), we can write:
\( \implies (\sqrt{3})^{n-1} = ((\sqrt{3})^2)^6 \)
\( \implies (\sqrt{3})^{n-1} = (\sqrt{3})^{12} \)
By comparing the exponents, we get:
\( \implies n - 1 = 12 \)
\( \implies n = 12 + 1 \)
\( \implies n = 13 \)
Therefore, there are 13 terms in the G.P. Understanding how to convert powers between a base and its square root is key to solving such problems.
In simple words: Find the first term and the common ratio. Then, set the last term equal to the general formula \( ar^{n-1} \) and solve for \( n \) by matching the bases of the exponents.

🎯 Exam Tip: When the common ratio is a square root, convert the known term to the same base (the square root) before equating the exponents.

Question 9. Find the value of \( x + y + z \) if \( 1, x, y, z, 16 \) are in GP.
Answer:
The given sequence \( 1, x, y, z, 16 \) is a Geometric Progression (G.P.).
The first term is \( a = 1 \).
The fifth term is \( T_5 = 16 \).
Using the formula for the \( n \)th term \( T_n = ar^{n-1} \):
For \( T_5 \):
\( 16 = a \times r^{5-1} \)
\( 16 = 1 \times r^4 \)
\( 16 = r^4 \)
We know that \( 2^4 = 16 \).
So, \( r^4 = 2^4 \)
By comparing the bases, the common ratio is \( r = 2 \).
Now we can find \( x, y, \) and \( z \) using the common ratio:
Since \( \frac{x}{1} = r \), then \( x = 1 \times r = 1 \times 2 = 2 \).
Since \( \frac{y}{x} = r \), then \( y = x \times r = 2 \times 2 = 4 \).
Since \( \frac{z}{y} = r \), then \( z = y \times r = 4 \times 2 = 8 \).
So, \( x = 2, y = 4, \) and \( z = 8 \).
Finally, we need to find \( x + y + z \):
\( x + y + z = 2 + 4 + 8 = 14 \). In a geometric sequence, each term is simply the previous term multiplied by the common ratio.
In simple words: Find the common ratio using the first and last known terms. Then, multiply each term by this ratio to find the missing numbers, \( x, y, \) and \( z \). Add them up to get the final answer.

🎯 Exam Tip: For sequences with missing terms, it's often easiest to first find the common ratio (r) using the formula \( T_n = ar^{n-1} \) with the first and last known terms.

Question 10. The third term of a G.P. is 18 and its seventh term is \( 3\frac{5}{9} \). Find the tenth term of the G.P.
Answer:
In a Geometric Progression (G.P.), we are given:
The third term \( T_3 = 18 \).
The seventh term \( T_7 = 3\frac{5}{9} = \frac{3 \times 9 + 5}{9} = \frac{27+5}{9} = \frac{32}{9} \).
Let \( a \) be the first term and \( r \) be the common ratio. The formula for the \( n \)th term is \( T_n = ar^{n-1} \).
Using this, we write:
\( T_3 = ar^{3-1} = ar^2 = 18 \) (Equation i)
\( T_7 = ar^{7-1} = ar^6 = \frac{32}{9} \) (Equation ii)
To find \( r \), we divide Equation (ii) by Equation (i):
\( \implies \frac{ar^6}{ar^2} = \frac{\frac{32}{9}}{18} \)
\( \implies r^{6-2} = \frac{32}{9 \times 18} \)
\( \implies r^4 = \frac{32}{162} \)
Simplify the fraction:
\( \implies r^4 = \frac{16}{81} \)
We know that \( 16 = 2^4 \) and \( 81 = 3^4 \).
\( \implies r^4 = \left(\frac{2}{3}\right)^4 \)
By comparing the bases, the common ratio is \( r = \frac{2}{3} \).
Now, substitute \( r = \frac{2}{3} \) into Equation (i) to find \( a \):
\( ar^2 = 18 \)
\( a \left(\frac{2}{3}\right)^2 = 18 \)
\( a \left(\frac{4}{9}\right) = 18 \)
\( \implies a = \frac{18 \times 9}{4} \)
\( \implies a = \frac{9 \times 9}{2} \)
\( \implies a = \frac{81}{2} \)
So, the first term is \( a = \frac{81}{2} \) and the common ratio is \( r = \frac{2}{3} \).
Finally, we need to find the tenth term (\( T_{10} \)):
\( T_{10} = ar^{10-1} = ar^9 \)
\( T_{10} = \frac{81}{2} \times \left(\frac{2}{3}\right)^9 \)
We can write \( 81 \) as \( 3^4 \).
\( T_{10} = \frac{3^4}{2^1} \times \frac{2^9}{3^9} \)
Now, simplify the powers:
\( T_{10} = \frac{2^{9-1}}{3^{9-4}} \)
\( T_{10} = \frac{2^8}{3^5} \)
Calculate the values: \( 2^8 = 256 \) and \( 3^5 = 243 \).
\( T_{10} = \frac{256}{243} \)
The tenth term of the G.P. is \( \frac{256}{243} \). When dealing with fractions as ratios, it is often helpful to express all numbers as powers of their prime factors to simplify calculations.
In simple words: Use the given third and seventh terms to set up two equations. Divide them to find the common ratio \( r \). Then, use this \( r \) with one of the original equations to find the first term \( a \). Finally, use \( a \) and \( r \) in the general formula to calculate the tenth term.

🎯 Exam Tip: Always convert mixed fractions to improper fractions before using them in calculations to avoid errors.

Question 11. The 5th, 8th and 11th terms of a G.P. are P, Q and S respectively. Show that \( Q^2 = PS \).
Answer:
Let the first term of the Geometric Progression (G.P.) be \( a \) and the common ratio be \( r \).
The formula for the \( n \)th term is \( T_n = ar^{n-1} \).
We are given:
The 5th term \( T_5 = P \). Using the formula, \( P = ar^{5-1} = ar^4 \).
The 8th term \( T_8 = Q \). Using the formula, \( Q = ar^{8-1} = ar^7 \).
The 11th term \( T_{11} = S \). Using the formula, \( S = ar^{11-1} = ar^{10} \).
We need to show that \( Q^2 = PS \).
Let's calculate \( Q^2 \):
\( Q^2 = (ar^7)^2 \)
\( Q^2 = a^2 (r^7)^2 \)
\( Q^2 = a^2 r^{14} \)
Now, let's calculate \( PS \):
\( PS = (ar^4) \times (ar^{10}) \)
\( PS = a \times a \times r^4 \times r^{10} \)
\( PS = a^2 r^{4+10} \)
\( PS = a^2 r^{14} \)
Since \( Q^2 = a^2 r^{14} \) and \( PS = a^2 r^{14} \), we can conclude that \( Q^2 = PS \). This property, where the middle term squared equals the product of two terms equidistant from it, is characteristic of all geometric progressions.
In simple words: Write out the formulas for the 5th, 8th, and 11th terms using 'a' for the first term and 'r' for the common ratio. Then, calculate \( Q^2 \) and \( PS \) separately. You will find that both results are the same, proving the statement.

🎯 Exam Tip: This problem demonstrates a key property of geometric progressions. When terms are equidistant in a G.P., the middle term is the geometric mean of the outer terms.