Maharashtra Board Class 11 Maths Part 1 Chapter 4 Sequences and Series 4.4 Solutions

Question 1. Verify whether the following sequences are H.P.
(i) \( \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \dots \)
(ii) \( \frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{12}, \dots \)
(iii) \( \frac{1}{7}, \frac{1}{9}, \frac{1}{11}, \frac{1}{13}, \frac{1}{15}, \dots \)
Answer:
Solution:
(i) \( \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \dots \)
Here, the reciprocal sequence is 3, 5, 7, 9, ...
\( \therefore t_1 = 3, t_2 = 5, t_3 = 7, \dots \)
\( \therefore t_2 - t_1 = t_3 - t_2 = t_4 - t_3 = 2 \), constant
\( \therefore \) The reciprocal sequence is an A.P.
\( \therefore \) the given sequence is H.P.

(ii) \( \frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{12}, \dots \)
Here, the reciprocal sequence is 3, 6, 9, 12 ...
\( \therefore t_1 = 3, t_2 = 6, t_3 = 9, t_4 = 12, \dots \)
\( \therefore t_2 - t_1 = t_3 - t_2 = t_4 - t_3 = 3 \), constant
\( \therefore \) The reciprocal sequence is an A.P.
\( \therefore \) The given sequence is H.P.

(iii) \( \frac{1}{7}, \frac{1}{9}, \frac{1}{11}, \frac{1}{13}, \frac{1}{15}, \dots \)
Here, the reciprocal sequence is 7, 9, 11, 13, 15, ......
\( \therefore t_1 = 7, t_2 = 9, t_3 = 11, t_4 = 13, \dots \)
\( \therefore t_2 - t_1 = t_3 - t_2 = t_4 - t_3 = 2 \), constant
\( \therefore \) The reciprocal sequence is an A.P.
\( \therefore \) The given sequence is H.P.
In simple words: To verify if a sequence is a Harmonic Progression (H.P.), first find the reciprocal of each term. If this new sequence of reciprocals forms an Arithmetic Progression (A.P.) (meaning the difference between consecutive terms is constant), then the original sequence is an H.P.

🎯 Exam Tip: Remember that a sequence is an H.P. if and only if its reciprocal sequence is an A.P. Clearly show the calculation of the common difference for the reciprocal sequence to prove it's an A.P.

Question 2. Find the nth term and hence find the 8th term of the following H.P.s:
(i) \( \frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \dots \)
(ii) \( \frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \dots \)
(iii) \( \frac{1}{5}, \frac{1}{10}, \frac{1}{15}, \frac{1}{20}, \dots \)
Answer:
Solution:
i. \( \frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \dots \) are in H.P.
\( \therefore \) 2, 5, 8, 11, ... are in A.P.
\( \therefore a = 2, d = 3 \)
\( t_n = a + (n-1)d \)
\( = 2 + (n-1)(3) \)
\( = 3n-1 \)
\( \therefore \) nth term of H.P. is \( \frac{1}{3n-1} \)
\( \therefore \) 8th term of H.P. \( = \frac{1}{3(8)-1} = \frac{1}{23} \)

ii. \( \frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \dots \) are in H.P.
\( \therefore \) 4, 6, 8, 10, ... are in A.P.
\( \therefore a = 4, d = 2 \)
\( t_n = a + (n-1)d \)
\( = 4 + (n-1)(2) \)
\( = 2n + 2 \)
\( \therefore \) nth term of H.P. \( = \frac{1}{2n+2} \)
\( \therefore \) 8th term of H.P. \( = \frac{1}{2(8)+2} = \frac{1}{18} \)

iii. \( \frac{1}{5}, \frac{1}{10}, \frac{1}{15}, \frac{1}{20}, \dots \) are in H.P.
\( \therefore \) 5, 10, 15, 20, ... are in H.P.
\( \therefore a = 5, d = 5 \)
\( t_n = a + (n-1)d \)
\( = 5 + (n-1)(5) \)
\( = 5n \)
\( \therefore \) nth term of H.P. \( = \frac{1}{5n} \)
\( \therefore \) 8th term of H.P. \( = \frac{1}{5(8)} = \frac{1}{40} \)
In simple words: To find the nth term of an H.P., first find the nth term of its reciprocal A.P. using the formula \( t_n = a + (n-1)d \), then take the reciprocal of that result. To find the 8th term, substitute n=8 into the nth term formula.

🎯 Exam Tip: Clearly identify 'a' (first term) and 'd' (common difference) for the reciprocal A.P. before applying the nth term formula. Remember that the nth term of the H.P. is the reciprocal of the nth term of the A.P., not the other way around.

Question 3. Find A.M. of two positive numbers whose G.M. and H.M. are 4 and \( \frac{16}{5} \).
Answer:
Solution:
G.M. = 4, H.M. = \( \frac{16}{5} \)
\( \therefore (G.M.)^2 = (A.M.) (H.M.) \)
\( \therefore 16 = A.M. \times \frac{16}{5} \)
\( \therefore A.M. = 5 \)
In simple words: For two positive numbers, the square of their Geometric Mean (G.M.) is equal to the product of their Arithmetic Mean (A.M.) and Harmonic Mean (H.M.). Using this relationship, we can calculate the unknown A.M. when G.M. and H.M. are given.

🎯 Exam Tip: The relation \( G^2 = A \times H \) is crucial for these types of problems. Ensure you correctly substitute the given values and perform the algebraic manipulation to solve for the required mean.

Question 4. Find H.M. of two positive numbers whose A.M. and G.M. are \( \frac{15}{2} \) and 6.
Answer:
Solution:
A.M. = \( \frac{15}{2} \), G.M. = 6
Now, \( (G.M.)^2 = (A.M.) (H.M.) \)
\( \therefore 6^2 = \frac{15}{2} \times H.M. \)
\( \therefore H.M. = 36 \times \frac{2}{15} \)
\( \therefore H.M. = \frac{24}{5} \)
In simple words: Using the property that the square of the Geometric Mean (G.M.) equals the product of the Arithmetic Mean (A.M.) and Harmonic Mean (H.M.), we can find the H.M. by plugging in the given A.M. and G.M. values into the formula.

🎯 Exam Tip: Always write down the formula \( G^2 = A \times H \) before substituting values. Pay attention to fractions and ensure accurate multiplication and division to avoid calculation errors.

Question 5. Find G.M. of two positive numbers whose A.M. and H.M. are 75 and 48.
Answer:
Solution:
A.M. = 75, H.M. = 48
\( (G.M.)^2 = (A.M.) (H.M.) \)
\( \therefore (G.M.)^2 = 75 \times 48 \)
\( \therefore (G.M.)^2 = 25 \times 3 \times 16 \times 3 \)
\( \therefore (G.M.)^2 = 5^2 \times 4^2 \times 3^2 \)
\( \therefore G.M. = 5 \times 4 \times 3 \)
\( \therefore G.M. = 60 \)
In simple words: To find the Geometric Mean (G.M.) given the Arithmetic Mean (A.M.) and Harmonic Mean (H.M.), use the relationship that \( G^2 = A \times H \). Calculate the product of A.M. and H.M., then take the square root to find G.M.

🎯 Exam Tip: When calculating \( (G.M.)^2 \), factorize the numbers (e.g., 75 and 48) into their prime factors before multiplying. This simplifies taking the square root, making the calculation more efficient and less prone to errors.

Question 6. Insert two numbers between \( \frac{1}{7} \) and \( \frac{1}{13} \) so that the resulting sequence is a H.P.
Answer:
Solution:
Let the required numbers be \( \frac{1}{H_1} \) and \( \frac{1}{H_2} \).
\( \therefore \frac{1}{7}, \frac{1}{H_1}, \frac{1}{H_2}, \frac{1}{13} \) are in H.P.
\( \therefore 7, H_1, H_2 \) and 13 are in A.P.
\( \therefore t_1 = a = 7 \) and \( t_4 = a + 3d = 13 \)
\( \therefore 7 + 3d = 13 \)
\( \therefore 3d = 6 \)
\( d = 2 \)
\( \therefore H_1 = t_2 = a + d = 7 + 2 = 9 \)
and \( H_2 = t_3 = a + 2d = 7 + 2(2) = 11 \)
\( \therefore \frac{1}{9} \) and \( \frac{1}{11} \) are the required numbers to be inserted between \( \frac{1}{7} \) and \( \frac{1}{13} \) so that the resulting sequence is a H.P.
In simple words: To insert numbers to form an H.P., first consider their reciprocals which will form an A.P. Find the missing terms in this A.P. using the first and last terms, then take the reciprocals of these A.P. terms to get the H.P. terms.

🎯 Exam Tip: Convert the H.P. problem into an A.P. problem by taking reciprocals. Carefully calculate the common difference 'd' for the A.P. and then find the intermediate terms before converting back to H.P.

Question 7. Insert two numbers between 1 and -27 so that the resulting sequence is a G.P.
Answer:
Solution:
Let the required numbers be G\( _1 \) and G\( _2 \).
\( \therefore 1, G_1, G_2, -27 \) are in G.P.
\( \therefore t_1 = 1, t_2 = G_1, t_3 = G_2, t_4 = -27 \)
\( \therefore t_1 = a = 1 \)
\( t_n = ar^{n-1} \)
\( \therefore t_4 = (1) r^{4-1} \)
\( \therefore -27 = r^3 \)
\( \therefore r^3 = (-3)^3 \)
\( \therefore r = -3 \)
\( \therefore G_1 = t_2 = ar = 1(-3) = -3 \)
\( \therefore G_2 = t_3 = ar^2 = 1(-3)^2 = 9 \)
\( \therefore -3 \) and 9 are the required numbers to be inserted between 1 and -27 so that the resulting sequence is a G.P.
In simple words: To insert numbers to form a Geometric Progression (G.P.), identify the first term (a) and the total number of terms. Use the formula for the nth term of a G.P. \( (t_n = ar^{n-1}) \) to find the common ratio (r), then use 'a' and 'r' to calculate the intermediate terms.

🎯 Exam Tip: For inserting terms in a G.P., carefully determine the value of 'n' (total terms) and use the formula \( t_n = ar^{n-1} \) to find the common ratio 'r'. Be mindful of negative ratios, as they cause alternating signs in the sequence.

Question 8. Find two numbers whose A.M. exceeds their G.M. by \( \frac{1}{2} \) and their H.M. by \( \frac{25}{26} \).
Answer:
Solution:
Let a, b be the two numbers.
A \( = \frac{a+b}{2}, G = \sqrt{ab}, H = \frac{2ab}{a+b} \)
According to the given conditions,
A \( = G + \frac{1}{2} \implies G = A - \frac{1}{2} \)
A \( = H + \frac{25}{26} \implies H = A - \frac{25}{26} \)
Now, \( G^2 = AH \)
\( (A - \frac{1}{2})^2 = A(A - \frac{25}{26}) \)
\( A^2 - A + \frac{1}{4} = A^2 - \frac{25}{26}A \)
\( -A + \frac{25}{26}A = -\frac{1}{4} \)
\( -\frac{1}{26}A = -\frac{1}{4} \)
\( \therefore A = \frac{26}{4} = \frac{13}{2} \)
From G \( = A - \frac{1}{2} \)
G \( = \frac{13}{2} - \frac{1}{2} = \frac{12}{2} = 6 \)
We have A.M. \( = \frac{a+b}{2} = \frac{13}{2} \)
\( \therefore a+b = 13 \).....(iii)
and G.M. \( = \sqrt{ab} = 6 \)
\( \therefore ab = 36 \)
Substitute (iii) into \( ab=36 \):
\( a(13 - a) = 36 \)
\( 13a - a^2 = 36 \)
\( a^2 - 13a + 36 = 0 \)
\( (a-4)(a-9) = 0 \)
\( \therefore a = 4 \) or \( a = 9 \)
When a = 4, b = 13 - 4 = 9
When a = 9, b = 13 - 9 = 4
\( \therefore \) the two numbers are 4 and 9.
In simple words: This problem involves finding two numbers by using their relationships between A.M., G.M., and H.M. First, express G.M. and H.M. in terms of A.M. using the given conditions. Then, substitute these into the fundamental relation \( G^2 = AH \) to solve for A.M. Once A.M. is found, calculate G.M. and set up equations for \( a+b \) and \( ab \) to find the two numbers.

🎯 Exam Tip: Start by defining A, G, H in terms of a and b. Formulate equations from the given conditions: \( A = G + \frac{1}{2} \) and \( A = H + \frac{25}{26} \). Use the relation \( G^2 = AH \) to solve for A, then derive G, and finally solve the system of equations for a+b and ab to find 'a' and 'b'.

Question 9. Find two numbers whose A.M. exceeds G.M. by 7 and their H.M. by \( \frac{63}{5} \).
Answer:
Solution:
Let a, b be the two numbers.
A \( = \frac{a+b}{2}, G = \sqrt{ab}, H = \frac{2ab}{a+b} \)
According to the given conditions,
A \( = G + 7 \implies G = A - 7 \)
A \( = H + \frac{63}{5} \implies H = A - \frac{63}{5} \)
Now, \( G^2 = AH \)
\( (A-7)^2 = A (A-\frac{63}{5}) \)
\( A^2 - 14A + 49 = A^2 - \frac{63A}{5} \)
\( -14A + 49 = -\frac{63A}{5} \)
\( 49 = 14A - \frac{63A}{5} \)
\( 49 = \frac{70A - 63A}{5} \)
\( 49 = \frac{7A}{5} \)
\( \therefore A = \frac{49 \times 5}{7} = 7 \times 5 = 35 \)
\( \therefore A = 35 \)
Now, \( G = A - 7 = 35 - 7 = 28 \)
We know A.M. \( = \frac{a+b}{2} \)
\( \therefore \frac{a+b}{2} = 35 \)
\( \therefore a+b = 70 \).....(ii)
And G.M. \( = \sqrt{ab} \)
\( \therefore \sqrt{ab} = 28 \)
\( \therefore ab = 28^2 = 784 \)
Substitute (ii) into \( ab = 784 \):
\( a(70 - a) = 784 \)
\( 70a - a^2 = 784 \)
\( a^2 - 70a + 784 = 0 \)
\( a^2 - 56a - 14a + 784 = 0 \)
\( a(a - 56) - 14(a - 56) = 0 \)
\( (a - 56)(a - 14) = 0 \)
\( \therefore a = 14 \) or \( a = 56 \)
When a = 14, b = 70 - 14 = 56
When a = 56, b = 70 - 56 = 14
\( \therefore \) the two numbers are 14 and 56.
In simple words: Similar to the previous problem, this involves setting up equations for G.M. and H.M. in terms of A.M. from the given conditions. Then, use the \( G^2 = AH \) relation to find A.M., which in turn helps find G.M. Finally, solve the system of equations for \( a+b \) and \( ab \) to determine the two unknown numbers.

🎯 Exam Tip: This problem requires careful algebraic manipulation. Double-check your substitutions and calculations when solving the quadratic equation for 'A' and later for 'a'. The factorization step \( a^2 - 70a + 784 = 0 \) into \( (a-56)(a-14) = 0 \) is critical for finding the correct numbers.