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

Question 1. In a G.P., the fourth term is 48 and the eighth term is 768. Find the tenth term.

Answer: Solution:
Given, \(t_4 = 48\), \(t_8 = 768\)
\(t_n = ar^{n-1}\)
\(t_4 = ar^3\)
\(\therefore ar^3 = 48\) ...(i)
and \(ar^7 = 768\) ...(ii)
Equation (ii) \(\div\) equation (i), we get
\[ \frac{ar^7}{ar^3} = \frac{768}{48} \]
\(\implies r^4 = 16\)
\(\implies r = 2\)
Substituting \(r = 2\) in (i), we get
\(a.(2^3) = 48\)
\(a.8 = 48\)
\(\therefore a = 6\)
Now, \(t_{10} = ar^{10-1} = ar^9\)
\(t_{10} = 6 (2^9) = 6 \times 512\)
\(\therefore t_{10} = 3072\)
In simple words: This problem asks us to find a specific term (the tenth) in a geometric progression, given two other terms. We use the general formula for a G.P. term to find the common ratio and the first term, then calculate the required term.

🎯 Exam Tip: Pay close attention to setting up the equations correctly using the general G.P. formula \(t_n = ar^{n-1}\). Ensure accurate calculation of the common ratio and the first term, as these are foundational for finding any other term.

Question 2. For a G.P. a \( = \frac{4}{3} \) and \(t_7 = \frac{243}{1024}\), find the value of r.
Answer: Solution:
Given, \(a = \frac{4}{3}\), \(t_7 = \frac{243}{1024}\)
\(t_n = ar^{n-1}\)
\(t_7 = ar^6\)
\(\implies \frac{243}{1024} = \frac{4}{3} r^6\)
\(\implies r^6 = \frac{243}{1024} \times \frac{3}{4}\)
\(\implies r^6 = \frac{729}{4096}\)
\(\implies r^6 = \frac{3^6}{4^6}\)
\(\implies r = \frac{3}{4}\)
In simple words: We are given the first term 'a' and the seventh term 't7' of a Geometric Progression (G.P.). Using the formula \(t_n = ar^{n-1}\), we substitute the given values to solve for the common ratio 'r'.

🎯 Exam Tip: Remember the general formula for the \(n^{th}\) term of a G.P., \(t_n = ar^{n-1}\). Accurately substitute the given values for 'a' and 't_n' and solve for 'r' by isolating the term with the exponent.

Question 3. For a sequence, if \(t_n = \frac{5^{n-2}}{7^{n-3}}\), verify whether the sequence is a G.P. If it is a G.P., find its first term and the common ratio.
Answer: Solution:
The sequence (\(t_n\)) is a G.P., if \( \frac{t_n}{t_{n-1}} = \) constant, for all \(n \in N\).
Now, \(t_n = \frac{5^{n-2}}{7^{n-3}}\)
\(t_{n-1} = \frac{5^{(n-1)-2}}{7^{(n-1)-3}} = \frac{5^{n-3}}{7^{n-4}}\)
\[ \frac{t_n}{t_{n-1}} = \frac{5^{n-2}/7^{n-3}}{5^{n-3}/7^{n-4}} = \frac{5^{n-2}}{7^{n-3}} \times \frac{7^{n-4}}{5^{n-3}} \] \[ = \frac{5^{n-2}}{5^{n-3}} \times \frac{7^{n-4}}{7^{n-3}} \] \[ = 5^{(n-2)-(n-3)} \times 7^{(n-4)-(n-3)} \] \[ = 5^{n-2-n+3} \times 7^{n-4-n+3} \] \[ = 5^1 \times 7^{-1} \] \[ = \frac{5}{7} \] \(\therefore \frac{t_n}{t_{n-1}} = \frac{5}{7} = \) constant, for all \(n \in N\)
\(\therefore \) the sequence is a G.P. with common ratio \(r = \frac{5}{7}\)
First term \(t_1\) can be found by substituting \(n=1\) into the formula for \(t_n\):
\(t_1 = \frac{5^{1-2}}{7^{1-3}} = \frac{5^{-1}}{7^{-2}} = \frac{7^2}{5^1} = \frac{49}{5}\)
\(\therefore \) first term \(t_1 = \frac{49}{5}\)
In simple words: To verify if a sequence is a Geometric Progression (G.P.), we calculate the ratio of consecutive terms, \(t_n / t_{n-1}\). If this ratio is constant, it's a G.P., and that constant is the common ratio. Then, we find the first term by substituting n=1 into the given formula.

🎯 Exam Tip: For verifying a G.P., correctly computing the ratio \(t_n / t_{n-1}\) using exponent rules is crucial. If the ratio simplifies to a constant, state that it's a G.P. and clearly identify the common ratio and the first term.

Question 4. Find three numbers in G.P., such that their sum is 35 and their product is 1000.
Answer: Solution:
Let the three numbers in G.P. be \(\frac{a}{r}\), \(a\), \(ar\).
According to the first condition (product is 1000),
\[ \left(\frac{a}{r}\right) (a) (ar) = 1000 \] \[ \implies a^3 = 1000 \] \[ \implies a = 10 \] According to the second condition (sum is 35),
\[ \frac{a}{r} + a + ar = 35 \] \[ \implies a \left( \frac{1}{r} + 1 + r \right) = 35 \] Substituting the value of \(a=10\) into the sum equation,
\[ 10 \left( \frac{1}{r} + 1 + r \right) = 35 \] \[ \implies \frac{1}{r} + 1 + r = \frac{35}{10} \] \[ \implies \frac{1}{r} + r = \frac{35}{10} - 1 \] \[ \implies \frac{1}{r} + r = \frac{35-10}{10} = \frac{25}{10} \] \[ \implies \frac{1}{r} + r = \frac{5}{2} \] Multiply by \(2r\) to clear the denominators:
\[ 2 + 2r^2 = 5r \] \[ \implies 2r^2 - 5r + 2 = 0 \] Factor the quadratic equation:
\[ 2r^2 - 4r - r + 2 = 0 \] \[ \implies 2r(r - 2) - 1(r - 2) = 0 \] \[ \implies (2r - 1)(r - 2) = 0 \] This gives two possible values for \(r\):
\(\implies r = \frac{1}{2}\) or \(r = 2\)
Case 1: When \(r = \frac{1}{2}\) and \(a = 10\)
The numbers are: \( \frac{a}{r} = \frac{10}{1/2} = 20 \)
\( a = 10 \)
\( ar = 10 \left(\frac{1}{2}\right) = 5 \)
The three numbers are 20, 10, 5.
Case 2: When \(r = 2\) and \(a = 10\)
The numbers are: \( \frac{a}{r} = \frac{10}{2} = 5 \)
\( a = 10 \)
\( ar = 10 (2) = 20 \)
The three numbers are 5, 10, 20.
\(\therefore \) the three numbers in G.P. are 20, 10, 5 or 5, 10, 20.
In simple words: We assume three terms of a G.P. as a/r, a, ar. Using the given product, we find 'a'. Then, using the given sum, we form a quadratic equation in 'r' to find the common ratio. Finally, we substitute 'a' and 'r' to get the numbers.

🎯 Exam Tip: When dealing with an odd number of terms in a G.P. (like 3 or 5), use the form \(\frac{a}{r}, a, ar\) (or \(\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2\)) as it simplifies the product calculation significantly, directly yielding the value of 'a'.

Question 5. Find 4 numbers in G.P. such that the sum of the middle 2 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),
\[ \left(\frac{a}{r^3}\right) \left(\frac{a}{r}\right) (ar) (ar^3) = 1 \] \[ \implies a^4 = 1 \] \[ \implies a = 1 \quad \text{(since the terms are real)} \] According to the first condition (sum of middle 2 numbers is \( \frac{10}{3} \)),
\[ \frac{a}{r} + ar = \frac{10}{3} \] Substitute \(a=1\):
\[ \frac{1}{r} + r = \frac{10}{3} \] Multiply by \(3r\) to clear the denominators:
\[ 3 + 3r^2 = 10r \] \[ \implies 3r^2 - 10r + 3 = 0 \] Factor the quadratic equation:
\[ 3r^2 - 9r - r + 3 = 0 \] \[ \implies 3r(r - 3) - 1(r - 3) = 0 \] \[ \implies (3r - 1)(r - 3) = 0 \] This gives two possible values for \(r\):
\(\implies r = 3\) or \(r = \frac{1}{3}\)
Case 1: When \(r = 3\) and \(a = 1\)
The four numbers are:
\( \frac{a}{r^3} = \frac{1}{3^3} = \frac{1}{27} \)
\( \frac{a}{r} = \frac{1}{3} \)
\( ar = 1(3) = 3 \)
\( ar^3 = 1(3^3) = 27 \)
The numbers are \(\frac{1}{27}, \frac{1}{3}, 3, 27\).
Case 2: When \(r = \frac{1}{3}\) and \(a = 1\)
The four numbers are:
\( \frac{a}{r^3} = \frac{1}{(1/3)^3} = \frac{1}{1/27} = 27 \)
\( \frac{a}{r} = \frac{1}{1/3} = 3 \)
\( ar = 1 \left(\frac{1}{3}\right) = \frac{1}{3} \)
\( ar^3 = 1 \left(\frac{1}{3}\right)^3 = \frac{1}{27} \)
The numbers are \(27, 3, \frac{1}{3}, \frac{1}{27}\).
\(\therefore \) the four numbers in G.P. are \(\frac{1}{27}, \frac{1}{3}, 3, 27\) or \(27, 3, \frac{1}{3}, \frac{1}{27}\).
In simple words: We are looking for four numbers in a G.P. whose middle two terms sum to \(10/3\) and whose product is 1. We represent the terms in a way that simplifies the product to find the first term 'a' directly. Then, we use the sum condition to find the common ratio 'r' and finally determine the numbers.

🎯 Exam Tip: For an even number of terms in a G.P. (like 4), use the symmetric form \(\frac{a}{r^3}, \frac{a}{r}, ar, ar^3\). This simplifies the product calculation, allowing you to quickly solve for 'a', which is crucial for solving for 'r' efficiently.

Question 6. Find five numbers in G.P. such that their product is 243 and the sum of the second and fourth numbers is 10.
Answer: Solution:
Let the five numbers in G.P. be \(\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2\).
According to the first condition (product is 243),
\[ \left(\frac{a}{r^2}\right) \left(\frac{a}{r}\right) (a) (ar) (ar^2) = 243 \] \[ \implies a^5 = 243 \] \[ \implies a = 3 \] According to the second condition (sum of second and fourth numbers is 10),
\[ \frac{a}{r} + ar = 10 \] Substitute \(a=3\):
\[ \frac{3}{r} + 3r = 10 \] Multiply by \(r\) to clear the denominator:
\[ 3 + 3r^2 = 10r \] \[ \implies 3r^2 - 10r + 3 = 0 \] Factor the quadratic equation:
\[ 3r^2 - 9r - r + 3 = 0 \] \[ \implies 3r(r - 3) - 1(r - 3) = 0 \] \[ \implies (3r - 1)(r - 3) = 0 \] This gives two possible values for \(r\):
\(\implies r = 3\) or \(r = \frac{1}{3}\)
Case 1: When \(r = 3\) and \(a = 3\)
The five numbers are:
\( \frac{a}{r^2} = \frac{3}{3^2} = \frac{3}{9} = \frac{1}{3} \)
\( \frac{a}{r} = \frac{3}{3} = 1 \)
\( a = 3 \)
\( ar = 3(3) = 9 \)
\( ar^2 = 3(3^2) = 3(9) = 27 \)
The numbers are \(\frac{1}{3}, 1, 3, 9, 27\).
Case 2: When \(r = \frac{1}{3}\) and \(a = 3\)
The five numbers are:
\( \frac{a}{r^2} = \frac{3}{(1/3)^2} = \frac{3}{1/9} = 27 \)
\( \frac{a}{r} = \frac{3}{1/3} = 9 \)
\( a = 3 \)
\( ar = 3 \left(\frac{1}{3}\right) = 1 \)
\( ar^2 = 3 \left(\frac{1}{3}\right)^2 = 3 \left(\frac{1}{9}\right) = \frac{1}{3} \)
The numbers are \(27, 9, 3, 1, \frac{1}{3}\).
\(\therefore \) the five numbers in G.P. are \(\frac{1}{3}, 1, 3, 9, 27\) or \(27, 9, 3, 1, \frac{1}{3}\).
In simple words: We are given the product of five numbers in a G.P. and the sum of its second and fourth terms. By setting up the terms symmetrically, the product simplifies to find 'a'. Then, using the sum condition, we solve a quadratic equation for 'r' and determine the numbers.

🎯 Exam Tip: For an odd number of terms in a G.P. (like 3 or 5), choosing the symmetric representation (e.g., \(\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2\)) simplifies the product calculation, making it easier to solve for 'a' first. Then, use the other conditions to find 'r'.

Question 7. For a sequence, \(S_n = 4(7^n - 1)\), verify whether the sequence is a G.P.
Answer: Solution:
Given, \(S_n = 4(7^n - 1)\)
Then, \(S_{n-1} = 4(7^{n-1} - 1)\)
We know that \(t_n = S_n - S_{n-1}\)
\(\implies t_n = 4(7^n - 1) - 4(7^{n-1} - 1)\)
\(\implies t_n = 4[ (7^n - 1) - (7^{n-1} - 1) ]\)
\(\implies t_n = 4[ 7^n - 1 - 7^{n-1} + 1 ]\)
\(\implies t_n = 4[ 7^n - 7^{n-1} ]\)
\(\implies t_n = 4[ 7^{n-1}(7^1 - 1) ]\)
\(\implies t_n = 4[ 7^{n-1}(6) ]\)
\(\implies t_n = 24 \cdot 7^{n-1}\)
Now, we need to find \(t_{n-1}\):
\(t_{n-1} = 24 \cdot 7^{(n-1)-1} = 24 \cdot 7^{n-2}\)
To verify if it's a G.P., we check the ratio \( \frac{t_n}{t_{n-1}} \):
\[ \frac{t_n}{t_{n-1}} = \frac{24 \cdot 7^{n-1}}{24 \cdot 7^{n-2}} \] \[ = 7^{(n-1)-(n-2)} \] \[ = 7^{n-1-n+2} \] \[ = 7^1 \] \[ = 7 \] Since \( \frac{t_n}{t_{n-1}} = 7 \), which is a constant for all \(n \in N\), the sequence is a G.P. with common ratio \(r=7\).
In simple words: To determine if a sequence is a G.P., we first find its nth term (\(t_n\)) by subtracting the sum of the first \(n-1\) terms (\(S_{n-1}\)) from the sum of the first n terms (\(S_n\)). Then, we calculate the ratio of consecutive terms (\(t_n / t_{n-1}\)). If this ratio is a constant, the sequence is a G.P.

🎯 Exam Tip: Remember the fundamental relationship \(t_n = S_n - S_{n-1}\) for finding the nth term of a sequence when the sum of n terms is given. Correctly applying exponent rules when calculating the ratio \(t_n / t_{n-1}\) is key to proving a G.P.

Question 8. Find \(2 + 22 + 222 + 2222 + \dots\) upto n terms.
Answer: Solution:
Let \(S_n = 2 + 22 + 222 + \dots\) upto n terms
Factor out 2:
\(S_n = 2(1 + 11 + 111 + \dots\) upto n terms)
Multiply and divide by 9:
\(S_n = \frac{2}{9}(9 + 99 + 999 + \dots\) upto n terms)
Express each term as a difference of powers of 10:
\(S_n = \frac{2}{9}[ (10 - 1) + (100 - 1) + (1000 - 1) + \dots\) upto n terms]
Separate the terms into two sums:
\(S_n = \frac{2}{9}[ (10 + 100 + 1000 + \dots\) upto n terms) \( - (1 + 1 + 1 + \dots\) n times)]
The first part \((10 + 100 + 1000 + \dots)\) is a G.P. with \(a = 10\) and \(r = \frac{100}{10} = 10\).
The sum of n terms of a G.P. is \(S_n = a \frac{r^n - 1}{r - 1}\).
So, \(10 + 100 + \dots\) upto n terms \( = 10 \frac{10^n - 1}{10 - 1} = \frac{10}{9}(10^n - 1)\).
The second part \((1 + 1 + 1 + \dots\) n times) is simply \(n\).
Substitute these back into the expression for \(S_n\):
\(S_n = \frac{2}{9} \left[ \frac{10}{9}(10^n - 1) - n \right]\)
\(S_n = \frac{2}{81}[10(10^n - 1) - 9n]\)
In simple words: To find the sum of this sequence, we first factor out 2. Then, we transform each term into the form \((10^k - 1)/9\) by multiplying and dividing by 9. This allows us to split the sum into a geometric progression of powers of 10 and a simple sum of 1s, which are then calculated separately.

🎯 Exam Tip: For sequences involving repeated digits (like 2, 22, 222), the key is to factor out the common digit and then transform the terms into \((10^k - 1)/9\). This method simplifies the problem into summing a G.P. and a series of constants.

Question 9. Find the nth term of the sequence 0.6, 0.66, 0.666, 0.6666,.....
Answer: Solution:
The sequence is 0.6, 0.66, 0.666, 0.6666, ......
We can write the terms as:
\(t_1 = 0.6 = \frac{6}{10}\)
\(t_2 = 0.66 = \frac{66}{100}\)
\(t_3 = 0.666 = \frac{666}{1000}\)
In general, the \(n^{th}\) term \(t_n\) is a number with \(n\) sixes after the decimal point.
\(t_n = \frac{6}{10} + \frac{6}{100} + \frac{6}{1000} + \dots\) (n terms)
This is a G.P. with first term \(a = 0.6 = \frac{6}{10}\) and common ratio \(r = \frac{0.06}{0.6} = 0.1 = \frac{1}{10}\).
The sum of the first \(n\) terms of a G.P. is \(S_n = a \frac{1 - r^n}{1 - r}\) (since \(|r| < 1\)).
\(\therefore t_n = 0.6 \frac{1 - (0.1)^n}{1 - 0.1}\)
\(t_n = 0.6 \frac{1 - (0.1)^n}{0.9}\)
\(t_n = \frac{6/10}{9/10} [1 - (0.1)^n]\)
\(t_n = \frac{6}{9} [1 - (0.1)^n]\)
\(t_n = \frac{2}{3} [1 - (0.1)^n]\)
Alternatively, using the pattern approach directly for each term as in Question 8:
\(t_n = 0.66\dots6\) (n times)
\(t_n = 6 \times (0.11\dots1)\) (n times)
\(t_n = 6 \times \frac{1}{9} \times (0.99\dots9)\) (n times)
\(t_n = \frac{6}{9} \times (1 - (0.1)^n)\)
\(t_n = \frac{2}{3} [1 - (0.1)^n]\)
In simple words: The nth term of this sequence is represented by 'n' repetitions of the digit '6' after the decimal point. We can express this as a sum of a geometric progression, or by manipulating it into the form \((1 - 10^{-n})/9\) multiplied by 6, to derive a compact formula for the nth term.

🎯 Exam Tip: For repeating decimal sequences like these, convert the terms to fractions (e.g., 0.6 = 6/10, 0.66 = 66/100). Then, apply the technique of factoring out the repeating digit (e.g., 6) and expressing the remaining part in terms of powers of 10, often involving the sum of a G.P. or the form \(\frac{1}{9}(10^n - 1)\).

Question 10. Find \( \sum_{r=1}^{n} (5r^2 + 4r - 3) \).
Answer: Solution:
We need to find \( \sum_{r=1}^{n} (5r^2 + 4r - 3) \)
Using the linearity property of summation, we can split this into three separate sums:
\[ = 5 \sum_{r=1}^{n} r^2 + 4 \sum_{r=1}^{n} r - 3 \sum_{r=1}^{n} 1 \] Now, we use the standard summation formulas:
\( \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \)
\( \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \)
\( \sum_{r=1}^{n} 1 = n \)
Substitute these formulas:
\[ = 5 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 4 \left[ \frac{n(n+1)}{2} \right] - 3n \] To combine these, find a common denominator, which is 6:
\[ = \frac{5n(n+1)(2n+1)}{6} + \frac{12n(n+1)}{6} - \frac{18n}{6} \] Factor out \(\frac{n}{6}\):
\[ = \frac{n}{6} [5(n+1)(2n+1) + 12(n+1) - 18] \] Expand the terms inside the bracket:
\[ = \frac{n}{6} [5(2n^2 + 3n + 1) + 12n + 12 - 18] \] \[ = \frac{n}{6} [10n^2 + 15n + 5 + 12n + 12 - 18] \] Combine like terms:
\[ = \frac{n}{6} [10n^2 + (15n + 12n) + (5 + 12 - 18)] \] \[ = \frac{n}{6} [10n^2 + 27n - 1] \] In simple words: This problem asks for the sum of a polynomial expression from r=1 to n. We use the linearity of summation to split the expression into individual sums of powers of 'r', and then apply standard formulas for \( \sum r^2 \), \( \sum r \), and \( \sum 1 \). Finally, we simplify the resulting algebraic expression.

🎯 Exam Tip: Memorize the standard summation formulas for \( \sum r \), \( \sum r^2 \), and \( \sum r^3 \). When simplifying the final expression, always look for common factors (like \(n/6\) in this case) to make the algebraic manipulation easier and to present the answer in a compact form.

Question 11. Find \( \sum_{r=1}^{n} r(r - 3)(r - 2) \).
Answer: Solution:
We need to find \( \sum_{r=1}^{n} r(r - 3)(r - 2) \).
First, expand the product inside the summation:
\( r(r - 3)(r - 2) = r(r^2 - 2r - 3r + 6) \)
\( = r(r^2 - 5r + 6) \)
\( = r^3 - 5r^2 + 6r \)
Now, substitute this back into the summation:
\[ \sum_{r=1}^{n} (r^3 - 5r^2 + 6r) \] Using the linearity property of summation, we can split this into three separate sums:
\[ = \sum_{r=1}^{n} r^3 - 5 \sum_{r=1}^{n} r^2 + 6 \sum_{r=1}^{n} r \] Now, we use the standard summation formulas:
\( \sum_{r=1}^{n} r^3 = \left[ \frac{n(n+1)}{2} \right]^2 = \frac{n^2(n+1)^2}{4} \)
\( \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \)
\( \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \)
Substitute these formulas:
\[ = \frac{n^2(n+1)^2}{4} - 5 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 6 \left[ \frac{n(n+1)}{2} \right] \] Find a common denominator, which is 12:
\[ = \frac{3n^2(n+1)^2}{12} - \frac{10n(n+1)(2n+1)}{12} + \frac{36n(n+1)}{12} \] Factor out a common term \( \frac{n(n+1)}{12} \):
\[ = \frac{n(n+1)}{12} [3n(n+1) - 10(2n+1) + 36] \] Expand the terms inside the square bracket:
\[ = \frac{n(n+1)}{12} [3n^2 + 3n - (20n + 10) + 36] \] \[ = \frac{n(n+1)}{12} [3n^2 + 3n - 20n - 10 + 36] \] Combine like terms inside the bracket:
\[ = \frac{n(n+1)}{12} [3n^2 - 17n + 26] \] In simple words: First, expand the given product \(r(r-3)(r-2)\) to get a polynomial in 'r'. Then, apply the summation operator to each term of the polynomial separately, using the standard formulas for sums of powers of 'r'. Finally, simplify the resulting algebraic expression by finding a common denominator and factoring.

🎯 Exam Tip: Always expand the polynomial within the summation first to get terms in \(r^k\). Be meticulous with algebraic simplification, especially when finding a common denominator and factoring out common terms to arrive at the most concise form of the answer.

Question 12. Find \( \sum_{r=1}^{n} \frac{1^2+2^2+3^2+\dots+r^2}{2r+1} \).
Answer: Solution:
We know that the sum of the first \(r\) squares is \(1^2+2^2+3^2+\dots+r^2 = \frac{r(r+1)(2r+1)}{6}\).
Substitute this into the expression inside the summation:
\[ \frac{1^2+2^2+3^2+\dots+r^2}{2r+1} = \frac{\frac{r(r+1)(2r+1)}{6}}{2r+1} \] \[ = \frac{r(r+1)(2r+1)}{6(2r+1)} \] Cancel out the term \((2r+1)\) (assuming \(2r+1 \neq 0\), which is true for \(r \ge 1\)):
\[ = \frac{r(r+1)}{6} \] Now, we need to find the sum of this expression:
\[ \sum_{r=1}^{n} \frac{r(r+1)}{6} \] Factor out the constant \( \frac{1}{6} \):
\[ = \frac{1}{6} \sum_{r=1}^{n} r(r+1) \] Expand \(r(r+1)\):
\[ = \frac{1}{6} \sum_{r=1}^{n} (r^2 + r) \] Using the linearity of summation:
\[ = \frac{1}{6} \left[ \sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r \right] \] Apply the standard summation formulas:
\( \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \)
\( \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \)
Substitute these formulas:
\[ = \frac{1}{6} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right] \] Find a common denominator for the terms inside the bracket (which is 6):
\[ = \frac{1}{6} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6} \right] \] Factor out \( \frac{n(n+1)}{6} \) from the bracket:
\[ = \frac{1}{6} \left[ \frac{n(n+1)}{6} ( (2n+1) + 3 ) \right] \] \[ = \frac{1}{6} \left[ \frac{n(n+1)}{6} (2n+4) \right] \] \[ = \frac{n(n+1)(2n+4)}{36} \] Factor out 2 from \((2n+4)\):
\[ = \frac{n(n+1)2(n+2)}{36} \] \[ = \frac{2n(n+1)(n+2)}{36} \] Simplify the fraction:
\[ = \frac{n(n+1)(n+2)}{18} \] In simple words: This problem involves a summation where each term is a fraction with the sum of squares in the numerator and a linear term in the denominator. We first simplify the fraction using the known formula for the sum of squares. Then, we apply the summation formula for \( \sum (r^2 + r) \) to find the overall sum.

🎯 Exam Tip: When a summation involves complex expressions, always simplify the general term first. Recognizing the sum of squares formula and judiciously factoring out common terms during algebraic simplification are crucial steps for accuracy.

Question 13. Find \( \sum_{r=1}^{n} \frac{1^3+2^3+3^3+\dots+r^3}{(r+1)^2} \).
Answer: Solution:
We know that the sum of the first \(r\) cubes is \(1^3+2^3+3^3+\dots+r^3 = \left[ \frac{r(r+1)}{2} \right]^2 = \frac{r^2(r+1)^2}{4}\).
Substitute this into the expression inside the summation:
\[ \frac{1^3+2^3+3^3+\dots+r^3}{(r+1)^2} = \frac{\frac{r^2(r+1)^2}{4}}{(r+1)^2} \] Cancel out the term \((r+1)^2\) (assuming \(r+1 \neq 0\), which is true for \(r \ge 1\)):
\[ = \frac{r^2}{4} \] Now, we need to find the sum of this expression:
\[ \sum_{r=1}^{n} \frac{r^2}{4} \] Factor out the constant \( \frac{1}{4} \):
\[ = \frac{1}{4} \sum_{r=1}^{n} r^2 \] Apply the standard summation formula for \( \sum r^2 \):
\( \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \)
Substitute this formula:
\[ = \frac{1}{4} \left[ \frac{n(n+1)(2n+1)}{6} \right] \] \[ = \frac{n(n+1)(2n+1)}{24} \] In simple words: This problem involves a summation with the sum of cubes in the numerator and a squared term in the denominator. We simplify the fraction using the known formula for the sum of cubes. After simplification, the sum reduces to a simple sum of squares, which is then calculated using its standard formula.

🎯 Exam Tip: Promptly recognize and apply the formula for the sum of cubes: \( \sum r^3 = \left[ \frac{n(n+1)}{2} \right]^2 \). Simplifying the general term of the sum before applying the summation operator is a critical step that often reduces complex problems to standard formulas.

Question 14. Find \(2 \times 6 + 4 \times 9 + 6 \times 12 + \dots\) upto n terms.
Answer: Solution:
The given series is \(2 \times 6 + 4 \times 9 + 6 \times 12 + \dots\) upto n terms.
Observe the pattern in the first factors: 2, 4, 6, ...
This is an Arithmetic Progression (A.P.) with first term \(a_1 = 2\) and common difference \(d_1 = 2\).
The \(r^{th}\) term of this A.P. is \(t_r = a_1 + (r-1)d_1 = 2 + (r-1)2 = 2 + 2r - 2 = 2r\).
Observe the pattern in the second factors: 6, 9, 12, ...
This is an Arithmetic Progression (A.P.) with first term \(a_2 = 6\) and common difference \(d_2 = 3\).
The \(r^{th}\) term of this A.P. is \(t'_r = a_2 + (r-1)d_2 = 6 + (r-1)3 = 6 + 3r - 3 = 3r + 3\).
So, the \(r^{th}\) term of the given series is the product of these two \(r^{th}\) terms:
\(T_r = (2r)(3r+3)\)
\(T_r = 2r \cdot 3(r+1)\)
\(T_r = 6r(r+1) = 6r^2 + 6r\)
Now, we need to find the sum of this series upto n terms, \(S_n = \sum_{r=1}^{n} T_r\):
\[ S_n = \sum_{r=1}^{n} (6r^2 + 6r) \] Using the linearity property of summation:
\[ S_n = 6 \sum_{r=1}^{n} r^2 + 6 \sum_{r=1}^{n} r \] Apply the standard summation formulas:
\( \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \)
\( \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \)
Substitute these formulas:
\[ S_n = 6 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 6 \left[ \frac{n(n+1)}{2} \right] \] Simplify:
\[ S_n = n(n+1)(2n+1) + 3n(n+1) \] Factor out \(n(n+1)\):
\[ S_n = n(n+1) [ (2n+1) + 3 ] \] \[ S_n = n(n+1) [2n+4] \] Factor out 2 from \([2n+4]\):
\[ S_n = n(n+1) \cdot 2(n+2) \] \[ S_n = 2n(n+1)(n+2) \] In simple words: This problem asks for the sum of a series where each term is a product of two factors. We first identify that both sets of factors form separate Arithmetic Progressions. We then find the general (r-th) term for each A.P., multiply them to get the general term of the product series, and finally use summation formulas to find the total sum.

🎯 Exam Tip: For product series, always find the general term \(T_r\) first by analyzing the patterns of the individual factors. This typically involves identifying arithmetic or geometric progressions. Once \(T_r\) is a polynomial in \(r\), use the standard summation formulas to find \(S_n\).

Question 15. Find \(12^2 + 13^2 + 14^2 + 15^2 + \dots + 20^2\).
Answer: Solution:
The given sum is \(12^2 + 13^2 + 14^2 + 15^2 + \dots + 20^2\).
To find the sum of squares from an arbitrary starting point, we can express it as the difference of two sums from 1:
Sum \( = (1^2 + 2^2 + 3^2 + \dots + 20^2) - (1^2 + 2^2 + 3^2 + \dots + 11^2) \)
We use the standard summation formula for squares: \( \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \)
For the first sum (up to 20): \(n=20\)
\( S_{20} = \frac{20(20+1)(2 \times 20+1)}{6} = \frac{20 \times 21 \times 41}{6} \)
\( S_{20} = \frac{17220}{6} = 2870 \)
For the second sum (up to 11): \(n=11\)
\( S_{11} = \frac{11(11+1)(2 \times 11+1)}{6} = \frac{11 \times 12 \times 23}{6} \)
\( S_{11} = \frac{3036}{6} = 506 \)
Now, subtract the second sum from the first:
Total Sum \( = S_{20} - S_{11} = 2870 - 506 \)
Total Sum \( = 2364 \)
In simple words: To find the sum of squares starting from a number other than 1, we can use the technique of subtracting a partial sum from a full sum. We calculate the sum of squares from 1 up to the last term (20) and subtract the sum of squares from 1 up to the term just before the starting term (11).

🎯 Exam Tip: When calculating sums of squares (or cubes) for a range that doesn't start from 1, always use the formula \( \sum_{k=a}^{b} k^2 = \sum_{k=1}^{b} k^2 - \sum_{k=1}^{a-1} k^2 \). Ensure precise application of the summation formula and careful arithmetic, especially with the upper limits of the sums.

Question 16. Find \((50^2 - 49^2) + (48^2 - 47^2) + (46^2 - 45^2) + \dots + (2^2 - 1^2)\).
Answer: Solution:
The given series is \((50^2 - 49^2) + (48^2 - 47^2) + (46^2 - 45^2) + \dots + (2^2 - 1^2)\).
Each term is of the form \(a^2 - b^2\), which can be factored as \((a-b)(a+b)\).
For the first term: \(50^2 - 49^2 = (50-49)(50+49) = (1)(99) = 99\)
For the second term: \(48^2 - 47^2 = (48-47)(48+47) = (1)(95) = 95\)
For the third term: \(46^2 - 45^2 = (46-45)(46+45) = (1)(91) = 91\)
...
For the last term: \(2^2 - 1^2 = (2-1)(2+1) = (1)(3) = 3\)
So the series simplifies to: \(99 + 95 + 91 + \dots + 3\).
This is an Arithmetic Progression (A.P.) with first term \(a = 99\) and common difference \(d = 95 - 99 = -4\).
To find the sum, we first need to determine the number of terms, \(n\). The general term of an A.P. is \(t_n = a + (n-1)d\). The last term is 3.
\(3 = 99 + (n-1)(-4)\)
\(3 - 99 = -4(n-1)\)
\(-96 = -4(n-1)\)
\(24 = n-1\)
\(n = 25\)
So there are 25 terms in the series.
The sum of an A.P. is \(S_n = \frac{n}{2}(a + l)\), where \(l\) is the last term.
\(S_{25} = \frac{25}{2}(99 + 3)\)
\(S_{25} = \frac{25}{2}(102)\)
\(S_{25} = 25 \times 51\)
\(S_{25} = 1275\)
In simple words: This problem asks for the sum of a series composed of differences of squares. By using the algebraic identity \(a^2 - b^2 = (a-b)(a+b)\), each term simplifies to an odd number. The resulting series of odd numbers forms an Arithmetic Progression. We then find the number of terms and use the A.P. sum formula.

🎯 Exam Tip: Recognize the difference of squares factorization \(a^2 - b^2 = (a-b)(a+b)\) immediately for such problems. This simplification is key. Always identify the type of resulting sequence (A.P. or G.P.), then correctly find its first term, common difference/ratio, number of terms, and finally apply the appropriate sum formula.

Question 17. In a G.P., if \(t_2 = 7, t_4 = 1575\), find r.
Answer: Solution:
Given, \(t_2 = 7\), \(t_4 = 1575\)
The general term of a G.P. is \(t_n = ar^{n-1}\).
For \(t_2 = 7\):
\(ar^{2-1} = 7\)
\(\implies ar = 7\) ...(i)
For \(t_4 = 1575\):
\(ar^{4-1} = 1575\)
\(\implies ar^3 = 1575\) ...(ii)
Divide equation (ii) by equation (i):
\[ \frac{ar^3}{ar} = \frac{1575}{7} \] \[ \implies r^2 = 225 \] Taking the square root of both sides:
\[ r = \pm \sqrt{225} \] \[ \implies r = \pm 15 \] The common ratio \(r\) can be 15 or -15.
In simple words: We are given two terms of a Geometric Progression (G.P.), \(t_2\) and \(t_4\). Using the general formula for a G.P. term, \(t_n = ar^{n-1}\), we set up two equations. Dividing these equations allows us to eliminate 'a' and directly solve for the common ratio 'r'.

🎯 Exam Tip: When given two terms of a G.P., form two equations using \(t_n = ar^{n-1}\). Dividing the equation for the higher-indexed term by the equation for the lower-indexed term is an efficient way to find the common ratio 'r', as 'a' cancels out directly. Remember to consider both positive and negative roots when solving for \(r^2\).

Question 18. Find k so that k - 1, k, k + 2 are consecutive terms of a G.P.
Answer: Solution:
If k - 1, k, k + 2 are consecutive terms of a G.P., then the ratio of consecutive terms must be constant.
So, \( \frac{k}{k-1} = \frac{k+2}{k} \)
Cross-multiply:
\( k \times k = (k-1)(k+2) \)
\( k^2 = k^2 + 2k - k - 2 \)
\( k^2 = k^2 + k - 2 \)
Subtract \(k^2\) from both sides:
\( 0 = k - 2 \)
\(\implies k = 2\)
To verify, if \(k=2\), the terms are: \(2-1 = 1\), \(2\), \(2+2 = 4\).
The sequence is 1, 2, 4. The common ratio is \(2/1 = 2\) and \(4/2 = 2\). This is a G.P.
In simple words: For three terms to be in a Geometric Progression (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 set up this equality with the given expressions in terms of 'k' and solve the resulting algebraic equation to find the value of 'k'.

🎯 Exam Tip: The defining property of a G.P. is a constant common ratio. For three consecutive terms \(A, B, C\) in a G.P., the condition is \(B/A = C/B\), which simplifies to \(B^2 = AC\). Using this property directly can quickly lead to the solution for problems involving unknown terms in a G.P.

Question 19. If pth, qth and rth terms of a G.P. are x, y, z respectively, find the value of \(x^{q-r} \cdot y^{r-p} \cdot z^{p-q}\).
Answer: Solution:
Let \(A\) be the first term and \(R\) be the common ratio of the G.P.
The \(n^{th}\) term of a G.P. is \(t_n = A \cdot R^{n-1}\).
Given:
\(t_p = x \implies x = A \cdot R^{p-1}\) ...(1)
\(t_q = y \implies y = A \cdot R^{q-1}\) ...(2)
\(t_r = z \implies z = A \cdot R^{r-1}\) ...(3)
We need to find the value of \(x^{q-r} \cdot y^{r-p} \cdot z^{p-q}\).
Substitute the expressions for \(x, y, z\):
\[ = (A \cdot R^{p-1})^{q-r} \cdot (A \cdot R^{q-1})^{r-p} \cdot (A \cdot R^{r-1})^{p-q} \] Apply the power rules \((ab)^m = a^m b^m\) and \((a^x)^y = a^{xy}\):
\[ = A^{q-r} \cdot (R^{p-1})^{q-r} \cdot A^{r-p} \cdot (R^{q-1})^{r-p} \cdot A^{p-q} \cdot (R^{r-1})^{p-q} \] \[ = A^{q-r} \cdot R^{(p-1)(q-r)} \cdot A^{r-p} \cdot R^{(q-1)(r-p)} \cdot A^{p-q} \cdot R^{(r-1)(p-q)} \] Group terms with A and R separately:
\[ = A^{(q-r) + (r-p) + (p-q)} \cdot R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)} \] Now, calculate the exponent for A:
Exponent of A \( = q-r+r-p+p-q = 0 \)
So, \(A^0 = 1\).
Now, calculate the exponent for R:
Exponent of R \( = (p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q) \)
\( = (pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q) \)
Rearrange and sum the terms:
\( = pq - pr - q + r \) \( + qr - pq - r + p \) \( + pr - qr - p + q \) ------------------ \( = 0 \) So, \(R^0 = 1\).
Therefore, the value of the expression is \(A^0 \cdot R^0 = 1 \cdot 1 = 1\).
In simple words: This problem asks us to evaluate an expression involving terms of a G.P. raised to certain powers. We first define each term (x, y, z) using the general G.P. formula with a first term 'A' and common ratio 'R'. Substituting these definitions into the expression, we use exponent rules to simplify. The exponents for both 'A' and 'R' sum to zero, resulting in a final value of 1.

🎯 Exam Tip: For problems involving products of terms from a G.P. raised to different powers, express each term in the form \(AR^{n-1}\). Carefully apply exponent rules for multiplication and powers. The key is to sum the exponents for 'A' and 'R' separately; often, they will cancel out to zero, leading to a simple answer of 1.

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