GSEB Class 11 Statistics Solutions Chapter 9 Geometric Progression

Section - A

Question 1. Find the 6th term of a G.P. 0.2, 1, 5, ...
(a) 25
(b) 0.5
(c) 0.1
(d) 625
Answer: (d) 625
In simple words: To find the 6th term, identify the first term (a = 0.2) and the common ratio (r = 1/0.2 = 5). Then use the formula for the nth term of a G.P., \(T_n = a \cdot r^{n-1}\), with n = 6.

🎯 Exam Tip: Remember the formula for the nth term of a G.P., \(T_n = a \cdot r^{n-1}\), where 'a' is the first term and 'r' is the common ratio. Accuracy in calculating 'r' is crucial for correct results.

Question 2. If the first term of a G.P. is 'a' and the common ratio is 'b'. Find the (n + 1)th term,
(a) \(ab^n\)
(b) \(ar^n\)
Answer: (a) \(ab^n\)
In simple words: The general formula for the nth term of a G.P. is \(T_n = \text{first term} \cdot (\text{common ratio})^{n-1}\). If the first term is 'a' and the common ratio is 'b', then the \((n+1)\)th term will be \(a \cdot b^{((n+1)-1)}\), which simplifies to \(a \cdot b^n\).

🎯 Exam Tip: Pay close attention to the requested term index (n+1) vs. (n-1) in the exponent. Substitution of variables must be precise, matching the question's given symbols for first term and common ratio.

Question 3. For a G.P. 1, \( \sqrt{3} \), 3, \( 3\sqrt{3} \) find the 5th term?
(a) 9
(b) \(9\sqrt{3}\)
(c) 27
(d) \(\frac{(\sqrt{3})^{5}-1}{(\sqrt{3}-1)}\)
Answer: (a) 9
In simple words: Identify the first term (a = 1) and the common ratio (r = \( \sqrt{3} \)). To find the 5th term, apply the G.P. term formula \(T_n = a \cdot r^{n-1}\) where n = 5, which results in \(1 \cdot (\sqrt{3})^{5-1} = (\sqrt{3})^4 = 9\).

🎯 Exam Tip: Accurately calculate the common ratio. When dealing with roots in the common ratio, correctly apply exponent rules to simplify the expression for the required term.

Question 4. For a G.P., \(T_1 = a\) and \(T_5 = \frac{1}{a}\): where, \(a > 0\) then obtain Its third term.
(a) \(a^2\)
(b) 1
(c) \(\frac{(\sqrt{3})^{5}-1}{(\sqrt{3}-1)}\)
(d) \(a\)
Answer: (b) 1
In simple words: Given \(T_1 = a\) and \(T_5 = \frac{1}{a}\), we know \(T_n = T_1 \cdot r^{n-1}\). So, \(T_5 = a \cdot r^4 = \frac{1}{a}\), which means \(r^4 = \frac{1}{a^2}\). The third term \(T_3 = T_1 \cdot r^2 = a \cdot r^2\). Since \(r^2 = \sqrt{\frac{1}{a^2}} = \frac{1}{a}\), we find \(T_3 = a \cdot \frac{1}{a} = 1\).

🎯 Exam Tip: When given two terms of a G.P., find the common ratio 'r' first. The third term can be derived by applying the general G.P. term formula using the first term and the calculated common ratio.

Question 5. For a G.P. \(\frac{1}{9}\), \(\frac{1}{3}\), 1, ..., find the seventh term.
(a) 6561
(b) 243
(c) 81
(d) \(\frac{1}{81}\)
Answer: (c) 81
In simple words: The first term is \(a = \frac{1}{9}\) and the common ratio is \(r = \frac{1/3}{1/9} = 3\). Using the formula \(T_n = a \cdot r^{n-1}\) for the 7th term, we get \(T_7 = \frac{1}{9} \cdot 3^{7-1} = \frac{1}{9} \cdot 3^6 = \frac{1}{3^2} \cdot 3^6 = 3^{6-2} = 3^4 = 81\).

🎯 Exam Tip: Always determine the first term 'a' and the common ratio 'r' accurately from the given sequence. Be careful with fractional calculations for 'a' and 'r' to avoid errors in subsequent steps.

Question 6. Find the common ratio of a G.P. whose nth term is \(3(2^{n-1})\).
(a) 3
(b) 2
(c) 6
(d) 1
Answer: (b) 2
In simple words: The general form of the nth term of a G.P. is \(T_n = a \cdot r^{n-1}\). Comparing this with the given \(T_n = 3(2^{n-1})\), we can directly identify the common ratio 'r' as the base of the exponent \((n-1)\), which is 2. The first term 'a' is 3.

🎯 Exam Tip: When the nth term formula is provided, directly compare it to the standard \(T_n = a \cdot r^{n-1}\) form to identify the first term 'a' and the common ratio 'r'.

Question 7. For a G.P. 0.4, 0.04, 0.004, ....find the common ratio.
(a) 10
(b) 0.4
(c) 4
(d) 0.1
Answer: (d) 0.1
In simple words: The common ratio of a G.P. is found by dividing any term by its preceding term. For the given sequence, dividing the second term (0.04) by the first term (0.4) yields \(r = \frac{0.04}{0.4} = 0.1\).

🎯 Exam Tip: To find the common ratio 'r' in a G.P., divide the second term by the first term, or any term by its immediately preceding term. Be precise with decimal division.

Question 8. If x, 10, -25 are in G.P, then find the value of x.
(a) 4
(b) - 25
(c) - 4
(d) 2
Answer: (c) - 4
In simple words: In a geometric progression, the ratio of consecutive terms is constant. Therefore, \(\frac{10}{x} = \frac{-25}{10}\). Cross-multiplying gives \(100 = -25x\), so \(x = \frac{100}{-25} = -4\).

🎯 Exam Tip: The fundamental property of a G.P. is that the ratio of any two consecutive terms is constant. Set up the proportion correctly to solve for the unknown variable.

Question 9. The common ratio of a G.P. is -1 and its first term is -1, then find the sum of the first six terms of the G.P.
(a) 0
(c) 1
(d) 6
Answer: (a) 0
In simple words: Given \(a = -1\) and \(r = -1\), the terms of the G.P. are -1, 1, -1, 1, -1, 1. When the common ratio is -1 and the number of terms is even, the sum of the terms will be 0 because positive and negative terms cancel each other out. Alternatively, use the sum formula \(S_n = \frac{a(1-r^n)}{1-r}\) for \(r \ne 1\), which gives \(S_6 = \frac{-1(1-(-1)^6)}{1-(-1)} = \frac{-1(1-1)}{2} = \frac{-1(0)}{2} = 0\).

🎯 Exam Tip: For a G.P. with a common ratio of -1, terms alternate in sign. If the number of terms is even, the sum will be zero. If the number of terms is odd, the sum will be equal to the first term.

Question 10. The common ratio of a G.P. is 1 and \(S_{10} = 40\), then find the first term.
(a) 0
(b) 10
(c) 4
(d) 400
Answer: (c) 4
In simple words: If the common ratio \(r = 1\), then all terms in the G.P. are the same as the first term, 'a'. The sum of 'n' terms is simply \(n \cdot a\). Given \(S_{10} = 40\) and \(n = 10\), we have \(10 \cdot a = 40\), which implies \(a = 4\).

🎯 Exam Tip: When the common ratio \(r = 1\), the sum of n terms of a G.P. simplifies to \(S_n = n \cdot a\). This special case is simpler than using the general sum formula.

Section - B

Question 1. What is the nth term of the G.P. ar, ar², ar³ ...?
Answer: The nth term of the G.P. ar, ar², ar³, ... is \(a \cdot r^n\).
In simple words: For a geometric progression starting with 'a', 'ar', 'ar²', the nth term is \(T_n = a \cdot r^{n-1}\). However, if the sequence begins with 'ar', 'ar²', 'ar³', the first term effectively becomes 'ar', so the nth term is \(ar \cdot r^{n-1} = ar^n\).

🎯 Exam Tip: Always be careful to identify the actual "first term" of the given sequence when applying the nth term formula, as it might not be the standard 'a'.

Question 2. Find the common ratio of the G.P. 0.1, 0.01, 0.001
Answer: The common ratio of the G.P. 0.1, 0.01, 0.001 is \(r = \frac{0.01}{0.1} = 0.1\).
In simple words: To find the common ratio, divide any term by its preceding term. Dividing the second term (0.01) by the first term (0.1) gives 0.1.

🎯 Exam Tip: Consistently divide a term by its previous term to find the common ratio. Ensure decimal calculations are precise.

Question 3. Find the sum of twenty terms of the G.P. 7, 7, 7,...
Answer: For the G.P. 7, 7, 7, ..., the first term is \(a = 7\), and the common ratio is \(r = 1\). With \(n = 20\), the sum \(S_n\) is given by \(n \cdot a\).
Therefore, \(S_{20} = 20 \times 7 = 140\).
In simple words: When all terms in a G.P. are identical, it means the common ratio is 1. In such a case, the sum of 'n' terms is simply 'n' times the first term.

🎯 Exam Tip: Recognize the special case where the common ratio 'r' is 1. In this scenario, the terms are constant, and the sum of 'n' terms is just \(n \times a\).

Question 4. If in a G.P. the nth term is given as \(T_n = 2^{n+1}\), find the common ratio.
Answer: The given nth term is \(T_n = 2^{n+1}\). The standard nth term for a G.P. is \(T_n = a \cdot r^{n-1}\). To find the common ratio 'r', we can compare \(T_n = 2^{n+1}\) with \(a \cdot r^{n-1}\) or calculate the ratio of consecutive terms like \(\frac{T_2}{T_1}\).
\(T_1 = 2^{1+1} = 2^2 = 4\).
\(T_2 = 2^{2+1} = 2^3 = 8\).
The common ratio \(r = \frac{T_2}{T_1} = \frac{8}{4} = 2\).
In simple words: The common ratio of a geometric progression can be found by evaluating the first two terms from the given \(T_n\) formula and then dividing the second term by the first.

🎯 Exam Tip: An alternative method to find 'r' from \(T_n\) is to express \(T_n\) in the form \(a \cdot r^{n-1}\). Here, \(2^{n+1} = 2^2 \cdot 2^{n-1} = 4 \cdot 2^{n-1}\), directly showing \(a=4\) and \(r=2\).

Question 5. The numbers 4, 1, y are in G.P. Find the value of y.
Answer: Given that the numbers 4, 1, y are in G.P., the common ratio must be constant.
Therefore, \(\frac{1}{4} = \frac{y}{1}\).
This implies \(4y = 1\).
Thus, \(y = \frac{1}{4}\).
In simple words: In a geometric progression, the ratio of any term to its preceding term is constant. By setting the ratio of the second to first term equal to the ratio of the third to second term, we can solve for 'y'.

🎯 Exam Tip: The defining characteristic of a G.P. is its constant common ratio. Use this property to set up an equation and solve for any unknown term within the sequence.

Question 6. For a G.P., sum of any two consecutive terms is zero, then what will be the common ratio?
Answer: If the sum of any two consecutive terms is 0, let the terms be \(T_1\) and \(T_2\).
So, \(T_1 + T_2 = 0\).
In a G.P., \(T_1 = a\) and \(T_2 = ar\).
Substituting these, \(a + ar = 0\).
Factoring out 'a', we get \(a(1 + r) = 0\).
Since 'a' (the first term) cannot be zero in a G.P., it must be that \(1 + r = 0\).
Therefore, the common ratio \(r = -1\).
In simple words: If two consecutive terms sum to zero, it means the second term is the negative of the first term, indicating a common ratio of -1.

🎯 Exam Tip: Understand that if the sum of consecutive terms is zero, the common ratio must be negative. Specifically, if \(a + ar = 0\), then \(r = -1\) (assuming \(a \neq 0\)).

Question 7. For a G.P., if \(S_7 = 15\) and \(S_6 = 11\), then find the seventh term of the G.P.
Answer: Given \(S_7 = 15\) and \(S_6 = 11\).
The nth term of a sequence can be found by subtracting the sum of \((n-1)\) terms from the sum of 'n' terms, i.e., \(T_n = S_n - S_{n-1}\).
To find the seventh term \(T_7\), we use \(T_7 = S_7 - S_6\).
\(T_7 = 15 - 11 = 4\).
In simple words: The nth term of any sequence is the difference between the sum of the first 'n' terms and the sum of the first '\(n-1\)' terms.

🎯 Exam Tip: Remember the general relation \(T_n = S_n - S_{n-1}\). This is a valuable tool for finding individual terms when only sums are provided, applicable to both Arithmetic and Geometric Progressions.

Question 8. State whether the statement "if a, b, c, d are in G.P., then ad = bc" is true or false.
Answer: Given that a, b, c, d are in G.P.
This implies that the common ratio is constant: \(\frac{b}{a} = \frac{c}{b} = \frac{d}{c}\).
From \(\frac{b}{a} = \frac{c}{b}\), we get \(b^2 = ac\).
From \(\frac{c}{b} = \frac{d}{c}\), we get \(c^2 = bd\).
Considering \(\frac{b}{a} = \frac{d}{c}\) (the ratio of the second term to the first equals the ratio of the fourth term to the third), we can cross-multiply to get \(bc = ad\).
Therefore, the statement "ad = bc" is true.
In simple words: For any four terms in a G.P., the product of the first and fourth term equals the product of the second and third term, due to the constant common ratio.

🎯 Exam Tip: The property \(ad=bc\) for terms a, b, c, d in G.P. is often used in problem-solving. This is equivalent to \(b/a = c/b = d/c\), where \(b^2=ac\), \(c^2=bd\), and \(bc=ad\) are all valid relations.

Question 9. State whether the statement "T₁ = S1' is true or false in G.P.
Answer: For a G.P., \(T_1\) represents the first term.
\(S_1\) represents the sum of the first term.
Since the sum of the first term is simply the first term itself, \(S_1 = T_1\).
Hence, the statement “\(T_1 = S_1\)” is true.
In simple words: The sum of the first term alone is, by definition, the first term itself.

🎯 Exam Tip: This is a fundamental definition. \(S_1\) always equals \(T_1\) for any sequence, not just a G.P. Understand basic definitions to avoid confusion in simple cases.

Section - C

Question 1. Define Geometric progression.
Answer: A geometric progression (G.P.) is a sequence of non-zero real numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. If 'a' is the first term and 'r' is the common ratio, the nth term is given by \(T_n = a \cdot r^{n-1}\) for an integer \(n \ge 1\).
In simple words: A geometric progression is a list of numbers where each number is found by multiplying the previous one by a constant value.

🎯 Exam Tip: Key elements of a G.P. definition include "non-zero real numbers," "multiplying by a fixed non-zero number," and the specific terms "first term" and "common ratio."

Question 2. Define Geometric series.
Answer: If the terms of a geometric progression, such as a, ar, \(ar^2\), ..., \(a \cdot r^{n-1}\), are added together, the resulting sum is called a geometric series. It is typically denoted by S, where \(S = a + ar + ar^2 + \dots + a \cdot r^{n-1}\).
In simple words: A geometric series is the sum of the terms of a geometric progression.

🎯 Exam Tip: Distinguish between a geometric progression (the sequence of terms) and a geometric series (the sum of those terms). The notation 'S' usually indicates a series.

Question 3. If in a G.P., common ratio is 1 and \(S_8 = 24\), find the first term of the G.P.
Answer: If the common ratio \(r = 1\) in a G.P., the sum of 'n' terms is given by \(S_n = n \cdot a\).
Here, \(S_8 = 24\) and \(n = 8\).
Substituting these values, \(8 \cdot a = 24\).
Therefore, \(a = \frac{24}{8} = 3\).
Hence, the first term of the G.P. is 3.
In simple words: When the common ratio is 1, all terms are equal to the first term. So, the sum is simply the number of terms multiplied by the first term.

🎯 Exam Tip: Always check for special conditions like \(r=1\) or \(r=-1\) as they simplify the sum or term calculations significantly.

Question 4. For a G.P., \(T_1 = 2\) and the product of the first three terms is 1000. Find the common ratio.
Answer: For a G.P., given \(T_1 = 2\), which means \(a = 2\).
The product of the first three terms is \(T_1 \times T_2 \times T_3 = 1000\).
In terms of 'a' and 'r', the terms are a, ar, \(ar^2\).
So, \(a \cdot ar \cdot ar^2 = 1000\).
This simplifies to \(a^3 r^3 = 1000\), or \((ar)^3 = 1000\).
Substituting \(a = 2\), we get \((2r)^3 = 1000\).
\(8r^3 = 1000\).
\(r^3 = \frac{1000}{8} = 125\).
Since \(125 = 5^3\), we have \(r^3 = 5^3\).
Therefore, the common ratio \(r = 5\).
In simple words: Use the first term and the product of the first three terms to form an equation involving the common ratio, then solve for 'r'.

🎯 Exam Tip: When given the product of terms, representing the terms as a, ar, \(ar^2\) (or \(a/r\), a, ar for an odd number of terms) simplifies the calculation significantly by eliminating 'a' or 'r' easily.

Question 5. For a G.P., \(a = 2\) and \(r = 3\); then find the sum of first four terms.
Answer: Given \(a = 2\), \(r = 3\), and \(n = 4\).
Since \(r > 1\), the sum of the first 'n' terms of a G.P. is given by \(S_n = \frac{a(r^n - 1)}{r - 1}\).
Substituting the values:
\(S_4 = \frac{2(3^4 - 1)}{3 - 1}\)
\(S_4 = \frac{2(81 - 1)}{2}\)
\(S_4 = \frac{2(80)}{2}\)
\(S_4 = 80\).
Hence, the sum of the first four terms of the G.P. is 80.
In simple words: Use the formula for the sum of 'n' terms of a G.P., substituting the given first term, common ratio, and number of terms.

🎯 Exam Tip: Choose the appropriate sum formula based on the common ratio (i.e., \(r > 1\) or \(r < 1\)) to avoid negative denominators, although both formulas yield the same result.

Question 7. For a G.P., \(a = \frac{4}{9}\) and \(r = \frac{-3}{2}\). Find \(T_3\).
Answer: Given the first term \(a = \frac{4}{9}\) and the common ratio \(r = \frac{-3}{2}\).
To find the third term, \(T_3\), we use the formula \(T_n = a \cdot r^{n-1}\).
For \(n = 3\):
\(T_3 = a \cdot r^{3-1} = a \cdot r^2\)
\(T_3 = \frac{4}{9} \cdot \left(\frac{-3}{2}\right)^2\)
\(T_3 = \frac{4}{9} \cdot \frac{9}{4}\)
\(T_3 = 1\).
Therefore, the third term of the G.P. is 1.
In simple words: The third term is calculated by multiplying the first term by the common ratio squared.

🎯 Exam Tip: Pay close attention to negative common ratios, as squaring them will result in a positive value. Correctly applying exponents is essential.

Question 8. If the common ratio of a geometric progression is 2, find the ratio of Its 7th and 3rd terms.
Answer: Given the common ratio \(r = 2\). We need to find the ratio \(\frac{T_7}{T_3}\).
The formula for the nth term of a G.P. is \(T_n = a \cdot r^{n-1}\).
So, \(T_7 = a \cdot r^{7-1} = a \cdot r^6\).
And \(T_3 = a \cdot r^{3-1} = a \cdot r^2\).
The ratio \(\frac{T_7}{T_3} = \frac{a \cdot r^6}{a \cdot r^2}\)
\(\frac{T_7}{T_3} = r^{6-2} = r^4\).
Substitute \(r = 2\):
\(\frac{T_7}{T_3} = 2^4 = 16\).
The ratio of the 7th and 3rd terms is 16.
In simple words: The ratio of two terms in a G.P. is the common ratio raised to the power of the difference in their term indices.

🎯 Exam Tip: When finding the ratio of two terms \(T_m\) and \(T_n\), the result is \(r^{m-n}\). This shortcut avoids needing the first term 'a' and simplifies calculations.

Question 9. Find the required term of the following sequence using sequence formula:
(i) 2, 10, 50, ... (6th term)
Answer: For the sequence 2, 10, 50, ..., we need to find \(T_6\).
The first term \(a = 2\).
The common ratio \(r = \frac{10}{2} = 5\).
The term to find is \(n = 6\).
Using the formula \(T_n = a \cdot r^{n-1}\):
\(T_6 = 2 \cdot 5^{6-1}\)
\(T_6 = 2 \cdot 5^5\)
\(T_6 = 2 \cdot 3125\)
\(T_6 = 6250\).
In simple words: Calculate the first term and common ratio, then use the nth term formula to find the 6th term.

🎯 Exam Tip: Ensure precise calculation of the common ratio and correct application of exponents, especially for higher terms where numbers can become large.

(ii) 100, 50, 25, ... (7th term)
Answer: For the sequence 100, 50, 25, ..., we need to find \(T_7\).
The first term \(a = 100\).
The common ratio \(r = \frac{50}{100} = \frac{1}{2}\).
The term to find is \(n = 7\).
Using the formula \(T_n = a \cdot r^{n-1}\):
\(T_7 = 100 \cdot \left(\frac{1}{2}\right)^{7-1}\)
\(T_7 = 100 \cdot \left(\frac{1}{2}\right)^6\)
\(T_7 = 100 \cdot \frac{1}{64}\)
\(T_7 = \frac{100}{64}\)
\(T_7 = \frac{25}{16}\).
In simple words: Determine the first term and common ratio, then apply the nth term formula to calculate the 7th term, simplifying the fraction.

🎯 Exam Tip: When the common ratio is a fraction, ensure you correctly raise both the numerator and denominator to the power of \((n-1)\). Simplify the final fraction to its lowest terms.

(iii) \(\frac{1}{3}\), \(\frac{2}{9}\), \(\frac{4}{27}\) ........ (8th term)
Answer: For the sequence \(\frac{1}{3}\), \(\frac{2}{9}\), \(\frac{4}{27}\), ..., we need to find \(T_8\).
The first term \(a = \frac{1}{3}\).
The common ratio \(r = \frac{2/9}{1/3} = \frac{2}{9} \times \frac{3}{1} = \frac{2}{3}\).
The term to find is \(n = 8\).
Using the formula \(T_n = a \cdot r^{n-1}\):
\(T_8 = \frac{1}{3} \cdot \left(\frac{2}{3}\right)^{8-1}\)
\(T_8 = \frac{1}{3} \cdot \left(\frac{2}{3}\right)^7\)
\(T_8 = \frac{1}{3} \cdot \frac{2^7}{3^7}\)
\(T_8 = \frac{128}{3^8}\)
\(T_8 = \frac{128}{6561}\).
In simple words: Identify the first term and common ratio from the given fractions. Then compute the 8th term using the nth term formula, carefully raising the fractional common ratio to the correct power.

🎯 Exam Tip: Be precise when calculating the common ratio of fractional terms. Remember to raise both the numerator and denominator of the common ratio to the power of \((n-1)\).

(iv) 2, \(2\sqrt{2}\), 4, ... (5th term)
Answer: For the sequence 2, \(2\sqrt{2}\), 4, ..., we need to find \(T_5\).
The first term \(a = 2\).
The common ratio \(r = \frac{2\sqrt{2}}{2} = \sqrt{2}\).
The term to find is \(n = 5\).
Using the formula \(T_n = a \cdot r^{n-1}\):
\(T_5 = 2 \cdot (\sqrt{2})^{5-1}\)
\(T_5 = 2 \cdot (\sqrt{2})^4\)
\(T_5 = 2 \cdot (2^{1/2})^4\)
\(T_5 = 2 \cdot 2^{4/2}\)
\(T_5 = 2 \cdot 2^2\)
\(T_5 = 2 \cdot 4\)
\(T_5 = 8\).
In simple words: Determine the first term and common ratio involving a square root. Then, apply the nth term formula to calculate the 5th term, simplifying the root raised to a power.

🎯 Exam Tip: When the common ratio involves a root, convert the root to a fractional exponent to simplify calculations involving powers. For example, \(\sqrt{2} = 2^{1/2}\).

Section - D

Question 1. If \(T_5 = 405\) and \(T_7 = 3645\); then find \(T_4\).
Answer: Given \(T_5 = 405\) and \(T_7 = 3645\). We need to find \(T_4\).
The formula for the nth term of a G.P. is \(T_n = a \cdot r^{n-1}\).
For \(T_5\): \(a \cdot r^4 = 405\) .......(1)
For \(T_7\): \(a \cdot r^6 = 3645\) .......(2)
Divide equation (2) by equation (1):
\(\frac{a \cdot r^6}{a \cdot r^4} = \frac{3645}{405}\)
\(r^2 = 9\)
Therefore, \(r = \pm 3\).

Substitute \(r = \pm 3\) into equation (1):
\(a \cdot (\pm 3)^4 = 405\)
\(a \cdot 81 = 405\)
\(a = \frac{405}{81} = 5\).

Now, find \(T_4\) using \(T_4 = a \cdot r^{4-1} = a \cdot r^3\).
Substitute \(a = 5\) and \(r = \pm 3\):
\(T_4 = 5 \cdot (\pm 3)^3\)
\(T_4 = 5 \cdot (\pm 27)\)
\(T_4 = \pm 135\).
In simple words: Use the given 5th and 7th terms to find the common ratio and the first term. Then, apply these values to calculate the 4th term. Remember to consider both positive and negative values for the common ratio.

🎯 Exam Tip: When solving for 'r' from \(r^2 = \text{positive number}\), remember that 'r' can be both positive and negative. Calculate 'a' and subsequent terms for both possibilities.

Question 2. Find \(T_5\) and \(S_4\) of the geometric progression If the first term is \(\frac{27}{16}\) and common ratio is \(\frac{2}{3}\).
Answer: Given the first term \(a = \frac{27}{16}\) and the common ratio \(r = \frac{2}{3}\). We need to find \(T_5\) and \(S_4\).

**To find \(T_5\):**
Using the formula \(T_n = a \cdot r^{n-1}\):
\(T_5 = \frac{27}{16} \cdot \left(\frac{2}{3}\right)^{5-1}\)
\(T_5 = \frac{27}{16} \cdot \left(\frac{2}{3}\right)^4\)
\(T_5 = \frac{27}{16} \cdot \frac{16}{81}\)
\(T_5 = \frac{27}{81} = \frac{1}{3}\).

**To find \(S_4\):**
Since \(r = \frac{2}{3} < 1\), use the sum formula \(S_n = \frac{a(1 - r^n)}{1 - r}\).
\(S_4 = \frac{\frac{27}{16} \left(1 - \left(\frac{2}{3}\right)^4\right)}{1 - \frac{2}{3}}\)
\(S_4 = \frac{\frac{27}{16} \left(1 - \frac{16}{81}\right)}{\frac{1}{3}}\)
\(S_4 = \frac{27}{16} \left(\frac{81 - 16}{81}\right) \cdot 3\)
\(S_4 = \frac{27}{16} \cdot \frac{65}{81} \cdot 3\)
\(S_4 = \frac{1}{16} \cdot \frac{65}{3} \cdot 3\)
\(S_4 = \frac{65}{16}\).
In simple words: Calculate the 5th term using the nth term formula and the sum of the first four terms using the appropriate sum formula for a G.P.

🎯 Exam Tip: Be meticulous with calculations involving fractions, especially when raising them to powers and performing subtraction or division. Simplify fractions at each possible step.

Question 3. For a given G.P., \(a = 4\) and \(T_5 = \frac{1}{4}\); then find \(T_7\).
Answer: Given \(a = 4\) and \(T_5 = \frac{1}{4}\). We need to find \(T_7\).
First, find the common ratio 'r' using \(T_5\).
\(T_n = a \cdot r^{n-1}\)
\(T_5 = a \cdot r^4\)
\(\frac{1}{4} = 4 \cdot r^4\)
\(r^4 = \frac{1}{16}\)
\(r = \pm \sqrt[4]{\frac{1}{16}} = \pm \frac{1}{2}\).

Now, find \(T_7\) using \(T_7 = a \cdot r^{7-1} = a \cdot r^6\).
\(T_7 = 4 \cdot \left(\pm \frac{1}{2}\right)^6\)
\(T_7 = 4 \cdot \frac{1}{64}\)
\(T_7 = \frac{4}{64} = \frac{1}{16}\).
In simple words: First, use the given first term and fifth term to determine the common ratio. Then, apply this common ratio and the first term to calculate the seventh term.

🎯 Exam Tip: Remember that \(r^4 = \text{positive number}\) implies 'r' can be positive or negative. However, raising 'r' to an even power (like 6 for \(T_7\)) will make the result positive regardless of 'r's sign.

Question 4. For a given G.P., if \(T_2 = 9\) and \(T_5 = 243\); then find \(S_4\).
Answer: Given \(T_2 = 9\) and \(T_5 = 243\). We need to find \(S_4\).
Using the formula \(T_n = a \cdot r^{n-1}\):
\(T_2 = a \cdot r = 9\) .......(1)
\(T_5 = a \cdot r^4 = 243\) .......(2)
Divide equation (2) by equation (1):
\(\frac{a \cdot r^4}{a \cdot r} = \frac{243}{9}\)
\(r^3 = 27\)
\(r = 3\).

Substitute \(r = 3\) into equation (1):
\(a \cdot 3 = 9\)
\(a = 3\).

Now, find \(S_4\) using the sum formula \(S_n = \frac{a(r^n - 1)}{r - 1}\) (since \(r = 3 > 1\)).
\(S_4 = \frac{3(3^4 - 1)}{3 - 1}\)
\(S_4 = \frac{3(81 - 1)}{2}\)
\(S_4 = \frac{3(80)}{2}\)
\(S_4 = 3 \times 40 = 120\).
In simple words: First, use the given 2nd and 5th terms to find the common ratio and the first term. Then, calculate the sum of the first four terms using the G.P. sum formula.

🎯 Exam Tip: When given any two terms of a G.P., divide the higher-indexed term by the lower-indexed term to find 'r' raised to the power of the difference in their indices. Then, use 'r' to find 'a'.

Question 5. The first term of a G.P., Is 10 and \(T_4 = 0.08\). Find sum of first three terms.
Answer: Given \(a = 10\) and \(T_4 = 0.08\). We need to find \(S_3\).
First, find the common ratio 'r' using \(T_4\).
\(T_n = a \cdot r^{n-1}\)
\(T_4 = a \cdot r^3\)
\(0.08 = 10 \cdot r^3\)
\(r^3 = \frac{0.08}{10} = 0.008\)
\(r^3 = (0.2)^3\)
\(r = 0.2\).

Now, find \(S_3\) using the sum formula \(S_n = \frac{a(1 - r^n)}{1 - r}\) (since \(r = 0.2 < 1\)).
\(S_3 = \frac{10(1 - (0.2)^3)}{1 - 0.2}\)
\(S_3 = \frac{10(1 - 0.008)}{0.8}\)
\(S_3 = \frac{10(0.992)}{0.8}\)
\(S_3 = \frac{9.92}{0.8}\)
\(S_3 = 12.4\).
In simple words: Use the first term and the fourth term to calculate the common ratio. Then, apply the first term and common ratio to find the sum of the first three terms.

🎯 Exam Tip: Convert decimals to fractions if it simplifies calculations, or maintain decimal precision throughout. Be careful with decimal point placement in powers and divisions.

Question 6. For a geometric progression, \(T_1^2 = T_2^2\) and \(T_3 = 64\). Write the sequence.
Answer: Given for a geometric progression: \(T_1^2 = T_2^2\) and \(T_3 = 64\). We need to write the sequence.
From \(T_1^2 = T_2^2\):
\((a)^2 = (ar)^2\)
\(a^2 = a^2 r^2\)
Since \(a \ne 0\), we can divide by \(a^2\):
\(1 = r^2\)
\(r = \pm 1\).

However, if \(r = 1\), then \(T_3 = a \cdot 1^2 = a\). So \(a = 64\). The sequence would be 64, 64, 64, ..., which is a G.P.
If \(r = -1\), then \(T_3 = a \cdot (-1)^2 = a\). So \(a = 64\). The sequence would be 64, -64, 64, ..., which is also a G.P.
The question as written with \(T_1^2 = T_2^2\) implies the relation between \(T_1\) and \(T_2\) should be considered carefully. If \(T_1^2 = T_2\), then \(a^2=ar\), so \(a=r\). Given \(T_3 = ar^2 = a \cdot a^2 = a^3 = 64\). Thus \(a=4\). Then \(r=4\).
Let's re-evaluate based on a common interpretation. If the squares are equal, the terms themselves could be equal or opposite: \(T_1 = \pm T_2\).
If \(T_1 = T_2\), then \(a = ar \implies r = 1\). If \(T_3 = 64\), then \(a = 64\). Sequence: 64, 64, 64, ...
If \(T_1 = -T_2\), then \(a = -ar \implies r = -1\). If \(T_3 = 64\), then \(a = 64\). Sequence: 64, -64, 64, ...

Looking at the provided solution steps, it seems the intention was \(T_1^2 = T_2\) and \(T_3 = ar^2 = 64\), or perhaps \(T_1 = a\), \(T_2 = ar\), \(T_3 = ar^2\), and the relationship \(T_1^2 = T_2\) from the previous question implies \(a^2 = ar \implies a = r\).
Then \(T_3 = ar^2 = a(a^2) = a^3 = 64 \implies a = 4\).
Since \(a = r\), then \(r = 4\).
The first term is 4 and the common ratio is 4.
Hence, the sequence obtained is 4, 16, 64, ...
In simple words: Assuming the condition \(T_1^2 = T_2\) from similar problem patterns, the first term 'a' and common ratio 'r' are found to be 4. This leads to the G.P. sequence 4, 16, 64.

🎯 Exam Tip: Carefully interpret symbolic relationships given in the problem statement. When a square of a term is related to another term, check if it leads to simple relations between 'a' and 'r' or to multiple possible sequences.

Question 7. If 5, m, 20, t are In geometric progression; find m and t.
Answer: Given that 5, m, 20, t are in G.P.
In a G.P., the ratio of consecutive terms is constant.
So, \(\frac{m}{5} = \frac{20}{m}\)
Cross-multiplying: \(m^2 = 5 \times 20 = 100\)
Therefore, \(m = \pm 10\).

Now, use the second ratio equality: \(\frac{20}{m} = \frac{t}{20}\)
Cross-multiplying: \(400 = mt\)

Case 1: If \(m = 10\)
\(400 = 10t \implies t = \frac{400}{10} = 40\).
The sequence is 5, 10, 20, 40.

Case 2: If \(m = -10\)
\(400 = -10t \implies t = \frac{400}{-10} = -40\).
The sequence is 5, -10, 20, -40.
Thus, \(m = \pm 10\) and \(t = \pm 40\) (with corresponding signs).
In simple words: Since the terms are in G.P., equate the common ratios to find 'm'. Then use 'm' to find 't', considering both positive and negative possibilities for 'm'.

🎯 Exam Tip: When solving for a variable squared, remember to consider both positive and negative roots. This often leads to multiple possible sets of terms for the G.P.

Question 8. For a given G.P., If \(a = 10\), \(r = 0.1\) and \(T_n = 0.01\); then find n.
Answer: Given \(a = 10\), \(r = 0.1\), and \(T_n = 0.01\). We need to find 'n'.
The formula for the nth term of a G.P. is \(T_n = a \cdot r^{n-1}\).
Substitute the given values:
\(0.01 = 10 \cdot (0.1)^{n-1}\)
Divide by 10:
\(\frac{0.01}{10} = (0.1)^{n-1}\)
\(0.001 = (0.1)^{n-1}\)
Express 0.001 as a power of 0.1:
\((0.1)^3 = (0.1)^{n-1}\)
Since the bases are equal, the exponents must be equal:
\(3 = n-1\)
\(n = 3 + 1\)
\(n = 4\).
In simple words: Use the nth term formula and substitute the known values. Then, by expressing both sides of the equation with the same base, equate the exponents to solve for 'n'.

🎯 Exam Tip: When dealing with decimals like 0.1, 0.01, 0.001, recognize that they are powers of 0.1 (\(0.1 = 10^{-1}\), \(0.01 = 10^{-2}\), etc.). This helps in equating exponents.

Question 9. For a given G.P., If \(a = 1\), \(r = 3\) and \(S_n = 121\); then find n.
Answer: Given \(a = 1\), \(r = 3\), and \(S_n = 121\). We need to find 'n'.
Since \(r = 3 > 1\), use the sum formula \(S_n = \frac{a(r^n - 1)}{r - 1}\).
Substitute the given values:
\(121 = \frac{1(3^n - 1)}{3 - 1}\)
\(121 = \frac{3^n - 1}{2}\)
Multiply by 2:
\(121 \times 2 = 3^n - 1\)
\(242 = 3^n - 1\)
Add 1 to both sides:
\(243 = 3^n\)
Express 243 as a power of 3: \(3^5 = 243\).
So, \(3^5 = 3^n\)
Since the bases are equal, the exponents must be equal:
\(n = 5\).
In simple words: Substitute the given values into the sum formula for a G.P., then solve the resulting exponential equation for 'n'.

🎯 Exam Tip: Recognize common powers of small integers (like 2, 3, 5) to quickly solve exponential equations. Knowing that \(3^5 = 243\) saves time.

Question 10. If \(S_n = \frac{2}{3}(2^n - 1)\), find \(T_4\).
Answer: Given the sum of 'n' terms \(S_n = \frac{2}{3}(2^n - 1)\). We need to find \(T_4\).
The nth term of a sequence can be found using the relation \(T_n = S_n - S_{n-1}\).
Therefore, \(T_4 = S_4 - S_3\).

First, find \(S_4\):
\(S_4 = \frac{2}{3}(2^4 - 1) = \frac{2}{3}(16 - 1) = \frac{2}{3}(15) = 10\).

Next, find \(S_3\):
\(S_3 = \frac{2}{3}(2^3 - 1) = \frac{2}{3}(8 - 1) = \frac{2}{3}(7) = \frac{14}{3}\).

Now, calculate \(T_4\):
\(T_4 = S_4 - S_3 = 10 - \frac{14}{3}\)
\(T_4 = \frac{30}{3} - \frac{14}{3} = \frac{16}{3}\).
In simple words: Use the formula \(T_n = S_n - S_{n-1}\) to find the 4th term. Calculate \(S_4\) and \(S_3\) from the given \(S_n\) formula and then subtract.

🎯 Exam Tip: This method \(T_n = S_n - S_{n-1}\) is universally applicable to find any term from its sum formula. Be careful with fractional arithmetic during subtraction.

Question 11. If \(S_n = 4(3^n - 1)\), find \(T_{n+1}\).
Answer: Given \(S_n = 4(3^n - 1)\). We need to find \(T_{n+1}\).
Using the relation \(T_k = S_k - S_{k-1}\), we can find \(T_{n+1}\) as \(S_{n+1} - S_n\).

First, find \(S_{n+1}\) by replacing 'n' with '\(n+1\)' in the \(S_n\) formula:
\(S_{n+1} = 4(3^{n+1} - 1)\).

Now, substitute \(S_{n+1}\) and \(S_n\) into the relation:
\(T_{n+1} = 4(3^{n+1} - 1) - 4(3^n - 1)\)
\(T_{n+1} = 4[(3^{n+1} - 1) - (3^n - 1)]\)
\(T_{n+1} = 4[3^{n+1} - 1 - 3^n + 1]\)
\(T_{n+1} = 4[3^{n+1} - 3^n]\)
Factor out \(3^n\):
\(T_{n+1} = 4[3^n \cdot 3^1 - 3^n]\)
\(T_{n+1} = 4[3^n(3 - 1)]\)
\(T_{n+1} = 4[3^n(2)]\)
\(T_{n+1} = 8 \cdot 3^n\).
In simple words: To find the \((n+1)\)th term, subtract the sum of 'n' terms from the sum of \((n+1)\) terms using the given formula for \(S_n\).

🎯 Exam Tip: When simplifying expressions with exponents like \(3^{n+1} - 3^n\), always factor out the smallest power (e.g., \(3^n\)) to simplify. This is a common algebraic step in sequence problems.

Question 12. Find the product of the first three terms of a G.P. whose second term is 5.
Answer: Let the first three terms of a G.P. be \(\frac{a}{r}\), \(a\), \(ar\).
Given that the second term is 5, so \(a = 5\).
The product of the first three terms is \(\left(\frac{a}{r}\right) \times a \times (ar)\).
Product = \(a^3\).
Substitute \(a = 5\):
Product = \(5^3 = 125\).
In simple words: Represent the terms as \(a/r\), \(a\), \(ar\). The product simplifies to \(a^3\). Given the second term 'a' is 5, the product is \(5^3\).

🎯 Exam Tip: When the product of an odd number of terms in a G.P. is required, represent the terms symmetrically (\(a/r, a, ar\), or \(a/r^2, a/r, a, ar, ar^2\), etc.) to simplify the product to a power of the middle term 'a'.

Question 13. How many terms of a geometric progression 2, 4, 8, 16, ... would add to 126?
Answer: Given the G.P.: 2, 4, 8, 16, ... and \(S_n = 126\). We need to find 'n'.
The first term \(a = 2\).
The common ratio \(r = \frac{4}{2} = 2\).
Since \(r = 2 > 1\), use the sum formula \(S_n = \frac{a(r^n - 1)}{r - 1}\).
Substitute the given values:
\(126 = \frac{2(2^n - 1)}{2 - 1}\)
\(126 = \frac{2(2^n - 1)}{1}\)
\(126 = 2(2^n - 1)\)
Divide by 2:
\(\frac{126}{2} = 2^n - 1\)
\(63 = 2^n - 1\)
Add 1 to both sides:
\(64 = 2^n\)
Express 64 as a power of 2: \(2^6 = 64\).
So, \(2^6 = 2^n\)
Since the bases are equal, the exponents must be equal:
\(n = 6\).
In simple words: Identify the first term and common ratio. Use the sum formula for a G.P. to set up an equation with 'n' as the unknown, then solve it by expressing both sides with the same base.

🎯 Exam Tip: Proficiency in powers of 2 (e.g., \(2^5 = 32\), \(2^6 = 64\), \(2^7 = 128\)) is helpful for quickly solving exponential equations involving base 2.

Question 14. If for a G.P., \(T_n = 324\), \(S_n = 484\) and \(r = 3\); find a and n.
Answer: Given \(T_n = 324\), \(S_n = 484\), and \(r = 3\). We need to find 'a' and 'n'.
Use the sum formula for a G.P. when \(r > 1\) in terms of \(T_n\): \(S_n = \frac{rT_n - a}{r - 1}\).
Substitute the given values:
\(484 = \frac{3(324) - a}{3 - 1}\)
\(484 = \frac{972 - a}{2}\)
Multiply by 2:
\(484 \times 2 = 972 - a\)
\(968 = 972 - a\)
\(a = 972 - 968\)
\(a = 4\).

Now that we have 'a', use the nth term formula \(T_n = a \cdot r^{n-1}\) to find 'n'.
\(324 = 4 \cdot (3)^{n-1}\)
Divide by 4:
\(\frac{324}{4} = 3^{n-1}\)
\(81 = 3^{n-1}\)
Express 81 as a power of 3: \(3^4 = 81\).
So, \(3^4 = 3^{n-1}\)
Since the bases are equal, the exponents must be equal:
\(4 = n-1\)
\(n = 4 + 1\)
\(n = 5\).
In simple words: First, use the alternative sum formula involving the last term to find the first term 'a'. Then, use the nth term formula along with 'a' and 'r' to find 'n'.

🎯 Exam Tip: The formula \(S_n = \frac{rT_n - a}{r - 1}\) is particularly useful when \(T_n\) (the last term) is known along with \(S_n\) and 'r', as it directly helps find 'a'. Remember this variation.

Question 15. Find the sum of required terms for the following sequence using series formula:

Question 15. (1) 4, 16, 64, ... (first 4 terms)
Answer: For the given geometric progression, the first term \(a\) is 4, and the common ratio \(r\) is also 4. We need to find the sum of the first four terms (\(n=4\)). Using the sum formula \(S_n = \frac{a(r^n - 1)}{r-1}\), we find \(S_4 = \frac{4(4^4 - 1)}{4-1} = \frac{4(256 - 1)}{3} = \frac{4 \times 255}{3} = 4 \times 85 = 340\).
In simple words: With a starting term of 4 and a common ratio of 4, the sum of the first four terms of this geometric sequence adds up to 340.

🎯 Exam Tip: Accurately identifying the first term and common ratio is crucial. Remember to use the correct sum formula for \(r > 1\) or \(r < 1\).

Question 15. (2) 2, 3, \(\frac{9}{2}\), ... (first 5 terms)
Answer: For the given geometric progression, with the first term \(a=2\) and common ratio \(r=\frac{3}{2}\), the sum of the first five terms (\(n=5\)) is determined. Applying the formula \(S_n = \frac{a(1-r^n)}{1-r}\) (or equivalently \(S_n = \frac{a(r^n-1)}{r-1}\) since \(r > 1\)), we get \(S_5 = \frac{2(1-(\frac{3}{2})^5)}{1-\frac{3}{2}} = \frac{2(1-\frac{243}{32})}{-\frac{1}{2}} = \frac{2(\frac{32-243}{32})}{-\frac{1}{2}} = \frac{2(-\frac{211}{32})}{-\frac{1}{2}} = \frac{-211/16}{-1/2} = \frac{211}{16} \times 2 = \frac{211}{8}\).
In simple words: Starting with 2 and a ratio of 3/2, the total when adding the first five terms of this series is 211/8.

🎯 Exam Tip: Pay close attention to fractions in calculations and ensure correct exponentiation before subtraction. Simplify fractions at each step to avoid errors.

Question 15. (3) 100, 20, 4, ... (first 5 terms)
Answer: To find the sum of the first five terms of this geometric progression, we identify the first term \(a=100\) and the common ratio \(r=\frac{20}{100}=\frac{1}{5}\). Since \(r < 1\), we use the sum formula \(S_n = \frac{a(1-r^n)}{1-r}\) for \(n=5\). This yields \(S_5 = \frac{100(1-(\frac{1}{5})^5)}{1-\frac{1}{5}} = \frac{100(1-\frac{1}{3125})}{\frac{4}{5}} = \frac{100(\frac{3125-1}{3125})}{\frac{4}{5}} = \frac{100 \times 3124 / 3125}{4/5} = 100 \times \frac{5}{4} \times \frac{3124}{3125} = 125 \times \frac{3124}{3125} = \frac{3124}{25} = 124.96\).
In simple words: For a sequence starting at 100 with a common ratio of 1/5, the sum of its first five terms is 124.96.

🎯 Exam Tip: For decreasing geometric progressions (\(r < 1\)), using the \(1-r^n\) form of the sum formula can simplify calculations by avoiding negative denominators.

Question 15. (4) \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\) ... (first 10 terms)
Answer: For the given geometric progression, the first term \(a = \frac{1}{2}\) and the common ratio \(r = \frac{1/4}{1/2} = \frac{1}{2}\). We need to find the sum of the first 10 terms (\(n=10\)). Using the formula \(S_n = \frac{a(1-r^n)}{1-r}\), we calculate \(S_{10} = \frac{\frac{1}{2}(1-(\frac{1}{2})^{10})}{1-\frac{1}{2}} = \frac{\frac{1}{2}(1-\frac{1}{1024})}{\frac{1}{2}} = 1 - \frac{1}{1024} = \frac{1024-1}{1024} = \frac{1023}{1024}\).
In simple words: With a first term of 1/2 and a common ratio of 1/2, the sum of the first 10 terms of this sequence is 1023/1024.

🎯 Exam Tip: Recognize when the numerator and denominator cancel out simplifying the sum formula. Be precise with powers of fractions.

Section - E

Question 1. If the three positive numbers k + 4, 4k-2 and 7k + 1 are in G.P., find k.
Answer: Given that the three positive numbers \(k+4\), \(4k-2\), and \(7k+1\) form a geometric progression, we know that the square of the middle term equals the product of the other two terms. Thus, \((4k-2)^2 = (k+4)(7k+1)\). Expanding this equation gives \(16k^2 - 16k + 4 = 7k^2 + k + 28k + 4\). Simplifying, we get \(16k^2 - 7k^2 - 16k - 29k + 4 - 4 = 0\), which further reduces to \(9k^2 - 45k = 0\). Factoring this equation as \(9k(k-5)=0\) leads to two possible values for \(k\): 0 or 5. As the problem specifies positive numbers in the sequence, \(k=0\) would result in \(4, -2, 1\), which is not a sequence of all positive numbers. Therefore, \(k=5\) is the valid solution.
In simple words: For three numbers to be in geometric progression, the middle term squared must equal the product of the other two. Solving this equation for \(k\) reveals that \(k\) must be 5 to maintain positive terms.

🎯 Exam Tip: Always verify solutions in the context of the problem (e.g., positive numbers). For quadratic equations, remember to check both roots.

Question 2. Find the maximum value of n such that the sum of the first n terms of a G.P. 1, 3, 3\(^2\), 3\(^3\) ... does not exceed 365.
Answer: We need to find the largest integer \(n\) such that the sum of the first \(n\) terms of the geometric progression \(1, 3, 3^2, \ldots\) does not exceed 365. Here, the first term \(a=1\) and the common ratio \(r=3\). The sum of \(n\) terms is given by \(S_n = \frac{a(r^n-1)}{r-1} = \frac{1(3^n-1)}{3-1} = \frac{3^n-1}{2}\). We set up the inequality \(\frac{3^n-1}{2} \le 365\), which simplifies to \(3^n-1 \le 730\), and further to \(3^n \le 731\). By evaluating powers of 3, we find \(3^5 = 243\), \(3^6 = 729\), and \(3^7 = 2187\). Since \(729 \le 731\) but \(2187 > 731\), the maximum possible value for \(n\) is 6.
In simple words: We are looking for how many terms of the sequence 1, 3, 9, ... can be added without exceeding a sum of 365. The maximum number of terms is 6.

🎯 Exam Tip: For inequalities involving exponents, test integer values of \(n\) to find the boundary condition. Ensure you choose the correct direction of inequality (maximum \(n\) vs. minimum \(n\)).

Question 3. Find the maximum value of n such that the sum of the first n terms of a G.P. 1, 2, 2\(^2\), 2\(^3\), ... is greater than or equal to 2000.
Answer: We aim to find the smallest integer \(n\) for which the sum of the first \(n\) terms of the geometric progression \(1, 2, 2^2, \ldots\) is greater than or equal to 2000. Here, the first term \(a=1\) and the common ratio \(r=2\). The sum of \(n\) terms is given by \(S_n = \frac{a(r^n-1)}{r-1} = \frac{1(2^n-1)}{2-1} = 2^n-1\). We set up the inequality \(2^n-1 \ge 2000\), which simplifies to \(2^n \ge 2001\). Examining powers of 2, we find \(2^{10} = 1024\) and \(2^{11} = 2048\). Since 2048 is the first power of 2 that meets or exceeds 2001, the minimum value for \(n\) is 11.
In simple words: We need to find the smallest number of terms in the sequence 1, 2, 4, ... that, when added together, result in a sum of 2000 or more. This requires at least 11 terms.

🎯 Exam Tip: When dealing with "greater than or equal to" for exponential growth, identify the first integer exponent that satisfies the condition.

Question 4. The sum of the first five terms of the G.P. y, \(\frac{y}{3}\), \(\frac{y}{9}\).... (where, y > 0) is 121. Find y.
Answer: For the given geometric progression where the first term is \(y\) and the common ratio is \(r = \frac{y/3}{y} = \frac{1}{3}\), the sum of the first five terms (\(S_5\)) is 121. Since the common ratio \(r < 1\), we use the sum formula \(S_n = \frac{a(1-r^n)}{1-r}\). Substituting the values, we get \(121 = \frac{y(1-(\frac{1}{3})^5)}{1-\frac{1}{3}}\). This simplifies to \(121 = \frac{y(1-\frac{1}{243})}{\frac{2}{3}} = \frac{y(\frac{243-1}{243})}{\frac{2}{3}}\). Solving for \(y\), we have \(121 = y \times \frac{242}{243} \times \frac{3}{2}\), which further simplifies to \(121 = y \times \frac{121}{81}\). Therefore, \(y = 81\).
In simple words: Given that the sum of the first five terms of a geometric progression starting with \(y\) and having a common ratio of 1/3 is 121, the initial term \(y\) is found to be 81.

🎯 Exam Tip: Be careful with algebraic manipulation, especially when dealing with fractions within the sum formula. Simplify terms progressively.

Question 5. For a geometric progression, S4 = 10 S2. Find the common ratio.
Answer: Given a geometric progression where the sum of the first four terms (\(S_4\)) is ten times the sum of the first two terms (\(S_2\)), we need to determine the common ratio (\(r\)). The formula for the sum of \(n\) terms is \(S_n = \frac{a(r^n-1)}{r-1}\). Substituting this into the given condition, we get \(\frac{a(r^4-1)}{r-1} = 10 \times \frac{a(r^2-1)}{r-1}\). Assuming \(a \ne 0\) and \(r \ne 1\), we can simplify this to \(r^4-1 = 10(r^2-1)\). Factoring the left side gives \((r^2-1)(r^2+1) = 10(r^2-1)\). If \(r^2-1 \ne 0\) (i.e., \(r \ne \pm 1\)), we can divide both sides by \((r^2-1)\), resulting in \(r^2+1 = 10\). This yields \(r^2 = 9\), so \(r = \pm 3\).
In simple words: When the sum of the first four terms of a geometric series is ten times the sum of the first two terms, the common ratio of the series must be either 3 or -3.

🎯 Exam Tip: When simplifying equations involving sum formulas, identify common factors like \((r-1)\) and \((r^2-1)\) to divide them out. Always consider edge cases like \(r=1\) or \(r=-1\) separately if division by zero could occur.

Question 6. For a geometric progression, the ratio of sum of the fifth and the third term to the difference of the fifth and the third term is 5:3. Find r.
Answer: We are given that for a geometric progression, the ratio of the sum of the fifth term (\(T_5\)) and the third term (\(T_3\)) to their difference is 5:3. Using the general term formula \(T_n = ar^{n-1}\), we have \(T_5 = ar^4\) and \(T_3 = ar^2\). The given condition translates to \(\frac{ar^4 + ar^2}{ar^4 - ar^2} = \frac{5}{3}\). Factoring out \(ar^2\) from the numerator and denominator, we get \(\frac{ar^2(r^2+1)}{ar^2(r^2-1)} = \frac{5}{3}\), which simplifies to \(\frac{r^2+1}{r^2-1} = \frac{5}{3}\). Cross-multiplying gives \(3(r^2+1) = 5(r^2-1)\), which expands to \(3r^2+3 = 5r^2-5\). Rearranging the terms, we find \(3+5 = 5r^2-3r^2\), leading to \(8 = 2r^2\), and finally \(r^2 = 4\). Thus, the common ratio \(r = \pm 2\).
In simple words: By setting up a ratio of the sum and difference of the fifth and third terms of a geometric progression, we can solve for the common ratio, which is found to be \( \pm 2 \).

🎯 Exam Tip: Remember to factor out common terms like \(ar^2\) to simplify complex fractional expressions involving terms of a GP. This makes solving for the common ratio much easier.

Question 7. The sum and the product of the three consecutive numbers in a geometric progression are 28 and 512 respectively. Find the three numbers of the G.P.
Answer: Let the three consecutive terms of the geometric progression be \(\frac{a}{r}\), \(a\), and \(ar\). The problem states their product is 512 and their sum is 28.
The product is \(\frac{a}{r} \times a \times ar = a^3\). So, \(a^3 = 512\), which implies \(a = 8\).
The sum is \(\frac{a}{r} + a + ar = 28\). Substituting \(a=8\), we get \(\frac{8}{r} + 8 + 8r = 28\).
Dividing by 8, we have \(\frac{1}{r} + 1 + r = \frac{28}{8} = \frac{7}{2}\).
Multiplying the entire equation by \(2r\) to clear denominators, we obtain \(2 + 2r + 2r^2 = 7r\). Rearranging gives the quadratic equation \(2r^2 - 5r + 2 = 0\).
Factoring this quadratic: \(2r^2 - 4r - r + 2 = 0 \implies 2r(r-2) - 1(r-2) = 0 \implies (r-2)(2r-1) = 0\).
This yields two possible values for the common ratio: \(r=2\) or \(r=\frac{1}{2}\).
If \(a=8\) and \(r=2\), the terms are \(\frac{8}{2}, 8, 8 \times 2\), which are 4, 8, 16.
If \(a=8\) and \(r=\frac{1}{2}\), the terms are \(\frac{8}{1/2}, 8, 8 \times \frac{1}{2}\), which are 16, 8, 4.
Both sets of terms satisfy the conditions.
In simple words: By representing the three terms of a G.P. as \(a/r, a, ar\), and using their given product and sum, we can solve for \(a\) and \(r\), leading to two possible sets of terms: 4, 8, 16 or 16, 8, 4.

🎯 Exam Tip: When the product of consecutive terms in a GP is given, setting terms as \(a/r, a, ar\) simplifies finding \(a\). Then use the sum to find \(r\).

Question 8. The sum and the product of the three consecutive terms of G.P. are 6 and -64 respectively. Find the three terms of the G.P.
Answer: Assume the three consecutive terms of the geometric progression are \(\frac{a}{r}\), \(a\), and \(ar\). We are given that their product is -64 and their sum is 6.
The product is \(\frac{a}{r} \times a \times ar = a^3\). So, \(a^3 = -64\), which means \(a = -4\).
The sum is \(\frac{a}{r} + a + ar = 6\). Substituting \(a=-4\), we get \(\frac{-4}{r} - 4 - 4r = 6\).
Rearranging the terms, we have \(\frac{-4}{r} - 4r = 10\). Multiplying by \(r\) yields \(-4 - 4r^2 = 10r\).
This simplifies to the quadratic equation \(4r^2 + 10r + 4 = 0\), or \(2r^2 + 5r + 2 = 0\) after dividing by 2.
Factoring the quadratic: \(2r^2 + 4r + r + 2 = 0 \implies 2r(r+2) + 1(r+2) = 0 \implies (r+2)(2r+1) = 0\).
This gives two possible common ratios: \(r=-2\) or \(r=-\frac{1}{2}\).
If \(a=-4\) and \(r=-2\), the terms are \(\frac{-4}{-2}, -4, -4 \times (-2)\), which are 2, -4, 8.
If \(a=-4\) and \(r=-\frac{1}{2}\), the terms are \(\frac{-4}{-1/2}, -4, -4 \times (-\frac{1}{2})\), which are 8, -4, 2.
Both sequences satisfy the given conditions.
In simple words: By using the generic form for three consecutive GP terms, the product helps find the middle term, and then the sum helps solve for the common ratio, yielding two possible sets of terms: 2, -4, 8 or 8, -4, 2.

🎯 Exam Tip: Remember that \(a^3 = -64\) implies \(a=-4\), not just \(4\), as negative numbers are allowed in geometric progressions unless specified otherwise. Factor quadratic equations carefully.

Question 9. A construction company offers a scheme on a flat to attract customers. In this scheme, customer has to pay Rs. 10,000 as the first installment and has to pay double the amount of the preceding installment in the subsequent annual installments. What is the total amount that the customer has to pay upto 10 installments?
Answer: In this construction scheme, the first installment is Rs. 10,000. Each subsequent installment is double the previous one, indicating a common ratio \(r=2\). We need to find the total amount payable for 10 installments (\(n=10\)). This is the sum of a geometric progression, calculated using the formula \(S_n = \frac{a(r^n - 1)}{r-1}\).
Substituting the values: \(S_{10} = \frac{\text{Rs. } 10,000(2^{10} - 1)}{2-1} = \text{Rs. } 10,000(1024 - 1) = \text{Rs. } 10,000 \times 1023 = \text{Rs. } 1,02,30,000\).
Therefore, the customer needs to pay a total of Rs. 1,02,30,000 for 10 installments.
In simple words: With a first installment of Rs. 10,000 and each subsequent installment doubling, the total payment for 10 installments will be Rs. 1,02,30,000.

🎯 Exam Tip: Recognize real-world problems as geometric progressions. Correctly identify the initial term, common ratio, and number of terms to apply the sum formula accurately.

Question 10. A banker counts 128 notes in the first minute and thereafter he counts half the number of notes he counted In the previous minute. How many notes he would count In five minutes?
Answer: A banker counts 128 notes in the first minute, and subsequently counts half the number of notes counted in the previous minute. This forms a geometric progression with the first term \(a=128\) and the common ratio \(r=\frac{1}{2}\). We need to find the total number of notes counted in five minutes (\(n=5\)), which is the sum of the first five terms (\(S_5\)).
Using the formula \(S_n = \frac{a(1-r^n)}{1-r}\), we have \(S_5 = \frac{128(1-(\frac{1}{2})^5)}{1-\frac{1}{2}} = \frac{128(1-\frac{1}{32})}{\frac{1}{2}} = 128 \times 2 \times (\frac{32-1}{32}) = 256 \times \frac{31}{32}\).
This calculation results in \(8 \times 31 = 248\) notes.
In simple words: Starting by counting 128 notes and then halving the count each minute, the banker will have counted a total of 248 notes after five minutes.

🎯 Exam Tip: Pay attention to keywords like "half the number" to correctly identify the common ratio. Ensure the sum formula for \(r < 1\) is applied correctly, including fraction arithmetic.

Question 11. Population of a village is 5000. Population increases at the rate of 2 % every year. What will be the population of the village after 10 years?
Answer: The initial population of the village is 5000. With an annual increase rate of 2%, the common ratio \(r\) for this geometric progression is \(1 + \frac{2}{100} = 1.02\). To find the population after 10 years, we are looking for the 11th term (\(T_{11}\)) in the sequence (as the initial population is \(T_1\), population after 1 year is \(T_2\), and so on).
Using the formula \(T_n = ar^{n-1}\), we calculate \(T_{11} = 5000(1.02)^{11-1} = 5000(1.02)^{10}\).
The value of \((1.02)^{10}\) can be computed using logarithms: \(\text{antilog}(10 \times \log(1.02)) = \text{antilog}(10 \times 0.0086) = \text{antilog}(0.086) = 1.219\).
Therefore, \(T_{11} = 5000 \times 1.219 = 6095\). The population after 10 years will be approximately 6095.
In simple words: With an initial population of 5000 and a 2% annual growth rate, the village's population will reach approximately 6095 after 10 years.

🎯 Exam Tip: For growth/depreciation problems over 'N' years, the term required is usually \(T_{N+1}\) (or \(ar^N\)). Logarithms are often needed for high powers of fractional or decimal common ratios.

Question 12. A car depreciates at the rate of 10 % every year. If the cost price of the car is Rs. 5,00,000, what will be the value of the car after 6 years?
Answer: The initial cost price of the car is Rs. 5,00,000. It depreciates at an annual rate of 10%, meaning the common ratio \(r\) for its value progression is \(1 - \frac{10}{100} = 0.9\). To find the car's value after 6 years, we need to calculate the 7th term (\(T_7\)) of this geometric progression (since the initial price is \(T_1\)).
Using the formula \(T_n = ar^{n-1}\), we get \(T_7 = \text{Rs. } 5,00,000(0.9)^{7-1} = \text{Rs. } 5,00,000(0.9)^6\).
Given \((0.9)^6 = 0.531441\), the value becomes \(T_7 = \text{Rs. } 5,00,000 \times 0.531441 = \text{Rs. } 2,65,720.50\).
Thus, the car's value after 6 years will be Rs. 2,65,720.50.
In simple words: A car bought for Rs. 5,00,000 loses 10% of its value each year. After 6 years, its value will be Rs. 2,65,720.50.

🎯 Exam Tip: Depreciation problems use \(r = 1 - \text{rate}\). Remember that "after N years" usually corresponds to the \((N+1)\)-th term or \(ar^N\) in the general term formula.