Maharashtra Board Class 11 Maths Part 1 Chapter 4 Sequences and Series 4.1 Solutions

Question 1. Verify whether the following sequences are G.P. If so, write t_n.
(i) 2, 6, 18, 54, .....
(ii) 1, -5, 25, -125, .....
(iii) \( \sqrt{5}, \frac{1}{\sqrt{5}}, \frac{1}{5\sqrt{5}}, \frac{1}{25\sqrt{5}}, \dots \)
(iv) 3, 4, 5, 6, .....
(v) 7, 14, 21, 28, .....
Answer:
(i) 2, 6, 18, 54, .....
\( t_1 = 2, t_2 = 6, t_3 = 18, t_4 = 54, \dots \)
Here, \( \frac{t_2}{t_1} = \frac{6}{2} = 3 \)
\( \frac{t_3}{t_2} = \frac{18}{6} = 3 \)
\( \frac{t_4}{t_3} = \frac{54}{18} = 3 \)
Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.
Here, \( a = 2, r = 3 \)
\( t_n = ar^{n-1} \)
\( \therefore t_n = 2(3^{n-1}) \)
(ii) 1, -5, 25, -125, .....
\( t_1 = 1, t_2 = -5, t_3 = 25, t_4 = -125, \dots \)
Here, \( \frac{t_2}{t_1} = \frac{-5}{1} = -5 \)
\( \frac{t_3}{t_2} = \frac{25}{-5} = -5 \)
\( \frac{t_4}{t_3} = \frac{-125}{25} = -5 \)
Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.
Here, \( a = 1, r = -5 \)
\( t_n = ar^{n-1} \)
\( \therefore t_n = (-5)^{n-1} \)
(iii) \( \sqrt{5}, \frac{1}{\sqrt{5}}, \frac{1}{5\sqrt{5}}, \frac{1}{25\sqrt{5}}, \dots \)
\( t_1 = \sqrt{5}, t_2 = \frac{1}{\sqrt{5}}, t_3 = \frac{1}{5\sqrt{5}}, t_4 = \frac{1}{25\sqrt{5}}, \dots \)
Here, \( \frac{t_2}{t_1} = \frac{1/\sqrt{5}}{\sqrt{5}} = \frac{1}{5} \)
\( \frac{t_3}{t_2} = \frac{1/(5\sqrt{5})}{1/\sqrt{5}} = \frac{1}{5} \)
\( \frac{t_4}{t_3} = \frac{1/(25\sqrt{5})}{1/(5\sqrt{5})} = \frac{1}{5} \)
Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.
Here, \( a = \sqrt{5}, r = \frac{1}{5} \)
\( t_n = ar^{n-1} \)
\( t_n = \sqrt{5} \left(\frac{1}{5}\right)^{n-1} = 5^{1/2} (5^{-1})^{n-1} = 5^{1/2} 5^{-(n-1)} = 5^{1/2 - (n-1)} = 5^{(3-2n)/2} \)
(iv) 3, 4, 5, 6,......
\( t_1 = 3, t_2 = 4, t_3 = 5, t_4 = 6, \dots \)
Here, \( \frac{t_2}{t_1} = \frac{4}{3} \)
\( \frac{t_3}{t_2} = \frac{5}{4} \)
\( \frac{t_4}{t_3} = \frac{6}{5} \)
Since, \( \frac{t_2}{t_1} \ne \frac{t_3}{t_2} \ne \frac{t_4}{t_3} \)
\( \therefore \) the given sequence is not a geometric progression.
(v) 7, 14, 21, 28, .....
\( t_1 = 7, t_2 = 14, t_3 = 21, t_4 = 28, \dots \)
Here, \( \frac{t_2}{t_1} = \frac{14}{7} = 2 \)
\( \frac{t_3}{t_2} = \frac{21}{14} = \frac{3}{2} \)
\( \frac{t_4}{t_3} = \frac{28}{21} = \frac{4}{3} \)
Since, \( \frac{t_2}{t_1} \ne \frac{t_3}{t_2} \ne \frac{t_4}{t_3} \)
\( \therefore \) the given sequence is not a geometric progression. In simple words: A sequence is a Geometric Progression (G.P.) if the ratio between any consecutive terms is constant. This constant is called the common ratio 'r'. If the ratio changes, it's not a G.P.

🎯 Exam Tip: Always calculate the ratio for at least three consecutive pairs to confirm if it's a G.P. For G.P., correctly identifying 'a' (first term) and 'r' (common ratio) is crucial for finding t_n.

Question 2. For the G.P.,
(i) if \( r = \frac{1}{3}, a = 9 \), find \( t_7 \).
(ii) if \( a = \frac{7}{243}, r = \frac{1}{3} \), find \( t_3 \).
(iii) if \( a = 7, r = -3 \), find \( t_6 \).
(iv) if \( a = \frac{2}{3}, t_6 = 162 \), find r.
Answer:
i. Given, \( r = \frac{1}{3}, a = 9 \)
\( t_n = ar^{n-1} \)
\( \therefore t_7 = 9 \times \left(\frac{1}{3}\right)^{7-1} = 9 \times \frac{1}{3^6} = \frac{9}{729} = \frac{1}{81} \)
ii. Given, \( a = \frac{7}{243}, r = \frac{1}{3} \)
\( t_n = ar^{n-1} \)
\( \therefore t_3 = \frac{7}{243} \times \left(\frac{1}{3}\right)^{3-1} = \frac{7}{243} \times \left(\frac{1}{3}\right)^2 = \frac{7}{243} \times \frac{1}{9} = \frac{7}{2187} \)
iii. Given, \( a = 7, r = -3 \)
\( t_n = ar^{n-1} \)
\( \therefore t_6 = 7 \times (-3)^{6-1} \)
\( = 7 \times (-3)^5 \)
\( = 7 \times (-243) = -1701 \)
iv. Given, \( a = \frac{2}{3}, t_6 = 162 \)
\( t_n = ar^{n-1} \)
\( \therefore t_6 = \frac{2}{3} (r^{6-1}) \)
\( 162 = \frac{2}{3} r^5 \)
\( \therefore r^5 = 162 \times \frac{3}{2} \)
\( r^5 = 81 \times 3 \)
\( r^5 = 243 \)
\( \therefore r^5 = 3^5 \)
\( \therefore r = 3 \) In simple words: The n-th term of a G.P. is found using the formula \( t_n = ar^{n-1} \), where 'a' is the first term and 'r' is the common ratio. We use this formula to find specific terms or missing values like 'r'.

🎯 Exam Tip: Remember to correctly substitute 'a', 'r', and 'n' into the \( t_n \) formula. Pay close attention to negative bases and their powers in calculations.

Question 3. Which term of the G. P. 5, 25, 125, 625, ..... is 510?
Answer:
Solution:
Here, \( t_1 = a = 5 \)
\( r = \frac{t_2}{t_1} = \frac{25}{5} = 5 \)
Let \( t_n = 5^{10} \)
We know, \( t_n = ar^{n-1} \)
\( \therefore 5^{10} = 5 \times 5^{n-1} \)
\( \therefore 5^{10} = 5^{1+n-1} \)
\( \therefore 5^{10} = 5^n \)
\( \therefore n = 10 \)
\( \therefore 5^{10} \) is the 10th term of the G.P. In simple words: To find which term of a G.P. a specific value is, we set the \( t_n \) formula equal to that value and solve for 'n', which represents the term number.

🎯 Exam Tip: When dealing with powers, simplify the base numbers to match each other to easily solve for the exponent 'n'.

Question 4. For what values of \( x, \frac{4}{3}, x, \frac{4}{27} \) are in G. P.?
Answer:
Solution:
Given sequence \( \frac{4}{3}, x, \frac{4}{27} \) are in geometric progression.
In a G.P., the ratio of consecutive terms is constant.
\( \therefore \frac{t_2}{t_1} = \frac{t_3}{t_2} \)
\( \therefore \frac{x}{4/3} = \frac{4/27}{x} \)
\( \therefore x^2 = \frac{4}{3} \times \frac{4}{27} \)
\( \therefore x^2 = \frac{16}{81} \)
\( \therefore x = \pm \sqrt{\frac{16}{81}} \)
\( \therefore x = \pm \frac{4}{9} \) In simple words: For three terms to be in a G.P., the ratio of the second to the first term must be equal to the ratio of the third to the second term. We use this property to find the unknown value.

🎯 Exam Tip: Remember that taking the square root can result in both positive and negative values for 'x'. Include both possibilities in your final answer.

Question 5. If for a sequence, \( t_n = \frac{5^{n-3}}{2^{n-3}} \), show that the sequence is a G. P. Find its first term and the common ratio.
Answer:
Solution:
The sequence \( (t_n) \) is a G.P., if \( \frac{t_n}{t_{n-1}} = \text{constant, for all } n \in N \)
Now, \( t_n = \frac{5^{n-3}}{2^{n-3}} \)
\( t_{n-1} = \frac{5^{(n-1)-3}}{2^{(n-1)-3}} = \frac{5^{n-4}}{2^{n-4}} \)
\( \therefore \frac{t_n}{t_{n-1}} = \frac{5^{n-3}}{2^{n-3}} \times \frac{2^{n-4}}{5^{n-4}} \)
\( = \frac{5^{n-3} \cdot 2^{n-4}}{2^{n-3} \cdot 5^{n-4}} \)
\( = \frac{5^{n-3 - (n-4)}}{2^{n-3 - (n-4)}} \)
\( = \frac{5^{n-3-n+4}}{2^{n-3-n+4}} \)
\( = \frac{5^1}{2^1} = \frac{5}{2} \), which is a constant, for all \( n \in N \).
\( \therefore \) the sequence is a G. P. with common ratio \( r = \frac{5}{2} \).
First term, \( t_1 = \frac{5^{1-3}}{2^{1-3}} = \frac{5^{-2}}{2^{-2}} = \frac{2^2}{5^2} = \frac{4}{25} \) In simple words: A sequence is a Geometric Progression if the ratio of any term to its preceding term is a constant (the common ratio). To find the first term, simply substitute n=1 into the given \( t_n \) formula.

🎯 Exam Tip: When proving a sequence is a G.P., explicitly show that the ratio \( \frac{t_n}{t_{n-1}} \) simplifies to a constant, which will be your common ratio 'r'.

Question 6. Find three numbers in G. P. such that their sum is 21 and sum of their squares is 189.
Answer:
Solution:
Let the three numbers in G. P. be \( \frac{a}{r}, a, ar \).
According to the first condition,
\( \frac{a}{r} + a + ar = 21 \)
\( a \left(\frac{1}{r} + 1 + r\right) = 21 \)
\( \frac{1}{r} + 1 + r = \frac{21}{a} \)
\( \therefore \frac{1}{r} + r = \frac{21}{a} - 1 \) ...(i)
According to the second condition,
\( \left(\frac{a}{r}\right)^2 + a^2 + (ar)^2 = 189 \)
\( a^2 \left(\frac{1}{r^2} + 1 + r^2\right) = 189 \)
\( \frac{1}{r^2} + 1 + r^2 = \frac{189}{a^2} \)
\( \therefore \frac{1}{r^2} + r^2 = \frac{189}{a^2} - 1 \) ...(ii)
On squaring equation (i), we get
\( \left(\frac{1}{r} + r\right)^2 = \left(\frac{21}{a} - 1\right)^2 \)
\( \frac{1}{r^2} + r^2 + 2 = \frac{441}{a^2} - \frac{42}{a} + 1 \)
Substituting the value of \( \left(\frac{1}{r^2} + r^2\right) \) from (ii) into the above equation:
\( \left(\frac{189}{a^2} - 1\right) + 2 = \frac{441}{a^2} - \frac{42}{a} + 1 \)
\( \therefore \frac{189}{a^2} + 1 = \frac{441}{a^2} - \frac{42}{a} + 1 \)
\( \therefore \frac{189}{a^2} = \frac{441}{a^2} - \frac{42}{a} \)
\( \therefore \frac{42}{a} = \frac{441}{a^2} - \frac{189}{a^2} \)
\( \therefore \frac{42}{a} = \frac{252}{a^2} \)
Multiply by \( a^2 \) on both sides (assuming \( a \ne 0 \)):
\( \therefore 42a = 252 \)
\( \therefore a = \frac{252}{42} = 6 \)
Substituting the value of a in (i), we get
\( \frac{1}{r} + r = \frac{21}{6} - 1 \)
\( \frac{1+r^2}{r} = \frac{7}{2} - 1 \)
\( \frac{1+r^2}{r} = \frac{5}{2} \)
\( 2(1+r^2) = 5r \)
\( 2 + 2r^2 = 5r \)
\( 2r^2 - 5r + 2 = 0 \)
\( 2r^2 - 4r - r + 2 = 0 \)
\( 2r(r-2) - 1(r-2) = 0 \)
\( (2r-1)(r-2) = 0 \)
\( \therefore r = \frac{1}{2} \text{ or } r = 2 \)
When \( a = 6, r = \frac{1}{2} \):
The numbers are \( \frac{a}{r} = \frac{6}{1/2} = 12 \), \( a = 6 \), \( ar = 6 \times \frac{1}{2} = 3 \).
When \( a = 6, r = 2 \):
The numbers are \( \frac{a}{r} = \frac{6}{2} = 3 \), \( a = 6 \), \( ar = 6 \times 2 = 12 \).
\( \therefore \) the three numbers are 12, 6, 3 or 3, 6, 12.
Check:
First condition:
12, 6, 3 are in G.P. with \( r = \frac{1}{2} \)
12 + 6 + 3 = 21 (Condition satisfied)
Second condition:
\( 12^2 + 6^2 + 3^2 = 144 + 36 + 9 = 189 \) (Condition satisfied)
Thus, both the conditions are satisfied. In simple words: When given conditions on the sum and sum of squares for three numbers in a G.P., represent the numbers as \( \frac{a}{r}, a, ar \). Formulate two equations based on the given conditions, then solve them simultaneously for 'a' and 'r' to find the numbers.

🎯 Exam Tip: Choosing \( \frac{a}{r}, a, ar \) simplifies calculations significantly, as 'r' terms cancel out in product-related conditions. Always verify your answers against both original conditions.

Question 7. Find four numbers in G. P. such that sum of the middle two numbers is \( \frac{10}{3} \) and their product is 1.
Answer:
Solution:
Let the four numbers in G.P. be \( \frac{a}{r^3}, \frac{a}{r}, ar, ar^3 \).
According to the second condition (Product is 1):
\( \left(\frac{a}{r^3}\right) \left(\frac{a}{r}\right) (ar) (ar^3) = 1 \)
\( \therefore a^4 = 1 \)
\( \therefore a = 1 \) (Since the numbers are real, a=1 is the only real solution for \( a^4=1 \) if we consider positive roots, or we consider \( a=\pm 1 \). If \( a=-1 \), the sum condition might not hold well with \( r=1/3 \) or \( r=3 \). Let's proceed with \( a=1 \) as it's the usual convention.)
According to the first condition (Sum of middle two is \( \frac{10}{3} \)):
\( \frac{a}{r} + ar = \frac{10}{3} \)
Substituting \( a=1 \):
\( \frac{1}{r} + 1r = \frac{10}{3} \)
\( \frac{1+r^2}{r} = \frac{10}{3} \)
\( 3(1+r^2) = 10r \)
\( 3 + 3r^2 = 10r \)
\( 3r^2 - 10r + 3 = 0 \)
Factorizing the quadratic equation:
\( 3r^2 - 9r - r + 3 = 0 \)
\( 3r(r-3) - 1(r-3) = 0 \)
\( (r-3)(3r-1) = 0 \)
\( \therefore r = 3 \text{ or } r = \frac{1}{3} \)
Case 1: When \( r = 3, a = 1 \)
The four numbers are:
\( \frac{a}{r^3} = \frac{1}{(3)^3} = \frac{1}{27} \)
\( \frac{a}{r} = \frac{1}{3} \)
\( ar = 1(3) = 3 \)
\( ar^3 = 1(3)^3 = 27 \)
Case 2: When \( r = \frac{1}{3}, a = 1 \)
The four numbers are:
\( \frac{a}{r^3} = \frac{1}{(1/3)^3} = 27 \)
\( \frac{a}{r} = \frac{1}{1/3} = 3 \)
\( ar = 1\left(\frac{1}{3}\right) = \frac{1}{3} \)
\( ar^3 = 1\left(\frac{1}{3}\right)^3 = \frac{1}{27} \)
The four numbers in G.P. are \( \frac{1}{27}, \frac{1}{3}, 3, 27 \text{ or } 27, 3, \frac{1}{3}, \frac{1}{27} \). In simple words: For problems involving conditions on an even number of terms in a G.P., choose terms like \( \frac{a}{r^3}, \frac{a}{r}, ar, ar^3 \). This form makes the product calculation simple, allowing you to find 'a' first, then use 'a' to find 'r' from other conditions.

🎯 Exam Tip: Selecting the correct representation for G.P. terms (e.g., \( \frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2 \) for 5 terms or \( \frac{a}{r^3}, \frac{a}{r}, ar, ar^3 \) for 4 terms) can greatly simplify algebraic steps, especially for product conditions.

Question 8. Find five numbers in G. P. such that their product is 1024 and the fifth term is square of the third term.
Answer:
Solution:
Let the five numbers in G. P. be \( \frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2 \).
According to the given conditions:
Product of the five numbers is 1024:
\( \left(\frac{a}{r^2}\right) \times \left(\frac{a}{r}\right) \times a \times ar \times ar^2 = 1024 \)
\( \therefore a^5 = 1024 \)
\( \therefore a^5 = 4^5 \)
\( \therefore a = 4 \) ...(i)
Also, the fifth term is the square of the third term:
\( t_5 = (t_3)^2 \)
\( ar^2 = a^2 \)
Since \( a \ne 0 \), we can divide by \( a \):
\( r^2 = a \)
Substitute \( a=4 \) from (i):
\( r^2 = 4 \)
\( \therefore r = \pm 2 \)
Case 1: When \( a = 4, r = 2 \)
The five numbers are:
\( \frac{a}{r^2} = \frac{4}{2^2} = \frac{4}{4} = 1 \)
\( \frac{a}{r} = \frac{4}{2} = 2 \)
\( a = 4 \)
\( ar = 4 \times 2 = 8 \)
\( ar^2 = 4 \times 2^2 = 16 \)
Case 2: When \( a = 4, r = -2 \)
The five numbers are:
\( \frac{a}{r^2} = \frac{4}{(-2)^2} = \frac{4}{4} = 1 \)
\( \frac{a}{r} = \frac{4}{-2} = -2 \)
\( a = 4 \)
\( ar = 4 \times (-2) = -8 \)
\( ar^2 = 4 \times (-2)^2 = 16 \)
\( \therefore \) the five numbers in G.P. are 1, 2, 4, 8, 16 or 1, -2, 4, -8, 16. In simple words: When finding five numbers in a G.P. with a product condition, representing them symmetrically as \( \frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2 \) simplifies the product to \( a^5 \). Then, use the second condition to find 'r' in terms of 'a', leading to the solution.

🎯 Exam Tip: For an odd number of terms (like 5), picking the middle term as 'a' and terms as powers of 'r' (e.g., \( a/r^2, a/r, a, ar, ar^2 \)) is efficient. Remember to consider both positive and negative values for 'r' if \( r^2 \) is involved.

Question 9. The fifth term of a G. P. is x, eighth term of the G. P. is y and eleventh term of the G. P. is z. Verify whether \( y^2 = xz \).
Answer:
Solution:
Given, \( t_5 = x, t_8 = y, t_{11} = z \)
Since, the formula for the n-th term of a G.P. is \( t_n = ar^{n-1} \)
\( t_5 = ar^{5-1} = ar^4 \)
\( t_8 = ar^{8-1} = ar^7 \)
\( t_{11} = ar^{11-1} = ar^{10} \)
Consider the Left Hand Side (L.H.S.):
L.H.S. \( = y^2 = (t_8)^2 = (ar^7)^2 = a^2r^{14} \)
Consider the Right Hand Side (R.H.S.):
R.H.S. \( = xz = t_5 \cdot t_{11} = (ar^4) \cdot (ar^{10}) = a^2r^{4+10} = a^2r^{14} \)
Since L.H.S. \( = a^2r^{14} \) and R.H.S. \( = a^2r^{14} \), we have:
\( \therefore \) L.H.S. = R.H.S.
\( \therefore y^2 = xz \) (Verified) In simple words: This problem demonstrates a property of G.P.s: if terms are equally spaced, the middle term's square equals the product of the terms equidistant from it. Here, \( t_8 \) is the middle term between \( t_5 \) and \( t_{11} \) (with a gap of 3 terms on each side).

🎯 Exam Tip: For verification problems, clearly state the known formulas, substitute the given terms, and show step-by-step how L.H.S. equals R.H.S.

Question 10. If p, q, r, s are in G. P., show that p + q, q + r, r + s are also in G.P.
Answer:
Solution:
Given that p, q, r, s are in G.P.
\( \therefore \frac{q}{p} = \frac{r}{q} = \frac{s}{r} \)
Let the common ratio be k.
\( \therefore \frac{q}{p} = \frac{r}{q} = \frac{s}{r} = k \)
From this, we can write:
\( q = pk \)
\( r = qk \)
\( s = rk \)
We have to prove that p + q, q + r, r + s are also in G.P.
This means we need to prove that the ratio of consecutive terms in this new sequence is constant.
i.e., we need to prove that \( \frac{q+r}{p+q} = \frac{r+s}{q+r} \)
Consider the Left Hand Side (L.H.S.):
L.H.S. \( = \frac{q+r}{p+q} \)
Substitute \( q=pk \) and \( r=qk \):
L.H.S. \( = \frac{pk+qk}{p+pk} = \frac{k(p+q)}{p(1+k)} \) (Mistake in OCR here, \( p+pk = p(1+k) \), \( pk+qk = k(p+q) \). Let's use \( q=pk, r=pk^2 \) approach for clarity.)
Let's re-evaluate using \( q=pk, r=pk^2, s=pk^3 \):
L.H.S. \( = \frac{q+r}{p+q} = \frac{pk+pk^2}{p+pk} = \frac{pk(1+k)}{p(1+k)} = k \)
Now, consider the Right Hand Side (R.H.S.):
R.H.S. \( = \frac{r+s}{q+r} \)
Substitute \( r=pk^2 \) and \( s=pk^3 \):
R.H.S. \( = \frac{pk^2+pk^3}{pk+pk^2} = \frac{pk^2(1+k)}{pk(1+k)} = k \)
Since L.H.S. = R.H.S. \( = k \), the common ratio for the new sequence is k.
\( \therefore \) p + q, q + r, r + s are in G.P. In simple words: If terms p, q, r, s are in a G.P. with a common ratio k, then q=pk, r=qk, s=rk. By substituting these relationships into the ratios \( \frac{q+r}{p+q} \) and \( \frac{r+s}{q+r} \), we can show that both ratios simplify to k, proving that p+q, q+r, r+s also form a G.P.

🎯 Exam Tip: For sequence proof questions, express all terms in relation to the first term and common ratio (e.g., \( t_n = ar^{n-1} \)) and then substitute these into the expression you need to prove. Look for common factors to simplify the ratios.