CBSE Class 10 Mathematics Arithmetic Progression Assignment Set 07

Question 1. The fourth term of an AP is 0. Prove that its \(25^{\text{th}}\) term is triple its \(11^{\text{th}}\) term.
Answer: Let the first term of the AP be \(a\) and the common difference be \(d\).
The \(n^{\text{th}}\) term of an AP is given by \(a_n = a + (n-1)d\).
Given that the fourth term is 0:
\(a_4 = 0\)
\(a + 3d = 0 \implies a = -3d\)
We need to prove that \(a_{25} = 3a_{11}\).
Consider the Left Hand Side (LHS):
\(a_{25} = a + 24d\)
Substitute \(a = -3d\):
\(LHS = -3d + 24d = 21d\)
Now consider the Right Hand Side (RHS):
\(3a_{11} = 3(a + 10d)\)
\(3a_{11} = 3(-3d + 10d) = 3(7d) = 21d\)
Since \(LHS = RHS = 21d\), the \(25^{\text{th}}\) term is triple its \(11^{\text{th}}\) term.
In simple words: Using the given condition \(a_4 = 0\), we expressed \(a\) in terms of \(d\) and verified that both sides of the equation yield \(21d\).

Exam Tip: Always write the general formula for the \(n^{\text{th}}\) term explicitly before substituting the given values to secure method marks.

Question 2. Find the \(20^{\text{th}}\) term from the end of the AP \(3, 8, 13, \dots, 253\).
Answer: The given AP is \(3, 8, 13, \dots, 253\).
Here, the first term \(a = 3\), common difference \(d = 8 - 3 = 5\), and the last term \(l = 253\).
The formula to find the \(n^{\text{th}}\) term from the end is:
\(a_n \text{ from end} = l - (n-1)d\)
Substitute \(l = 253\), \(n = 20\), and \(d = 5\):
\(a_{20} = 253 - (20-1)5\)
\(a_{20} = 253 - (19 \times 5)\)
\(a_{20} = 253 - 95\)
\(a_{20} = 158\)
In simple words: To find a term from the end, you can treat the last term as the starting term and reverse the sign of the common difference, or use the direct formula with the last term.

Exam Tip: Alternatively, you can reverse the entire AP to form a new AP where the first term is 253 and the common difference is \(-5\), then find the \(20^{\text{th}}\) term normally.

Question 3. If the \(p^{\text{th}}\), \(q^{\text{th}}\), and \(r^{\text{th}}\) term of an AP is \(x\), \(y\), and \(z\) respectively, show that \(x(q-r) + y(r-p) + z(p-q) = 0\).
Answer: Let the first term of the AP be \(A\) and the common difference be \(D\).
According to the given conditions:
\(x = A + (p-1)D\)
\(y = A + (q-1)D\)
\(z = A + (r-1)D\)
Substitute these values into the expression \(x(q-r) + y(r-p) + z(p-q)\):
\(= [A + (p-1)D](q-r) + [A + (q-1)D](r-p) + [A + (r-1)D](p-q)\)
Group the terms containing \(A\) and \(D\):
\(= A[(q-r) + (r-p) + (p-q)] + D[(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)]\)
Simplify the coefficient of \(A\):
\((q - r + r - p + p - q) = 0\)
Simplify the coefficient of \(D\):
\([pq - pr - q + r] + [qr - pq - r + p] + [rp - rq - p + q] = 0\)
Thus, the expression becomes:
\(= A(0) + D(0) = 0\)
Hence proved.
In simple words: Substitute the general term formulas into the given expression and expand. All terms will cancel out to give zero.

Exam Tip: Use capital letters \(A\) and \(D\) if lowercase letters \(a\) and \(d\) are already used in the question for other variables to avoid confusion.

Question 4. Find the sum of first 40 positive integers divisible by 6 also find the sum of first 20 positive integers divisible by 5 or 6.
Answer: Part 1: Sum of first 40 positive integers divisible by 6.
The AP is \(6, 12, 18, \dots\) up to 40 terms.
Here, \(a = 6\), \(d = 6\), and \(n = 40\).
The last term \(a_{40} = 6 + (40-1)6 = 6 + 39 \times 6 = 6 + 234 = 240\).
Using the sum formula \(S_n = \frac{n}{2}(a + l)\):
\(S_{40} = \frac{40}{2}(6 + 240)\)
\(S_{40} = 20 \times 246 = 4920\)

Part 2: Sum of first 20 positive integers divisible by 5 or 6.
The numbers divisible by both 5 and 6 are multiples of their LCM, which is 30.
The sequence of numbers divisible by 30 is \(30, 60, 90, \dots\) up to 20 terms.
Here, \(a = 30\), \(d = 30\), and \(n = 20\).
The last term \(a_{20} = 30 + (20-1)30 = 30 + 19 \times 30 = 30 + 570 = 600\).
\(S_{20} = \frac{20}{2}(30 + 600)\)
\(S_{20} = 10 \times 630 = 6300\)
In simple words: First identify the arithmetic progressions formed by the given conditions, find their last terms, and apply the sum formula.

Exam Tip: When dealing with divisibility by two numbers, remember that numbers divisible by both must be accounted for using their LCM if overlapping sets are considered.

Question 5. A man arranges to pay a debt of Rs. 3600 in 40 monthly instalments which are in an AP. When 30 instalments are paid he dies leaving one third of the debt unpaid. Find the value of the first instalment.
Answer: Let the first instalment be \(a\) and the common difference be \(d\).
Total debt = Rs. 3600, which is paid in 40 instalments.
\(S_{40} = \frac{40}{2}[2a + (40-1)d] = 3600\)
\(20[2a + 39d] = 3600\)
\(2a + 39d = 180 \quad \text{--- (1)}\)
When 30 instalments are paid, one third of the debt remains unpaid. Therefore, the amount paid in 30 instalments is two thirds of the total debt:
Amount paid = \(\frac{2}{3} \times 3600 = 2400\)
\(S_{30} = \frac{30}{2}[2a + (30-1)d] = 2400\)
\(15[2a + 29d] = 2400\)
\(2a + 29d = \frac{2400}{15} = 160 \quad \text{--- (2)}\)
Subtract equation (2) from equation (1):
\((2a + 39d) - (2a + 29d) = 180 - 160\)
\(10d = 20 \implies d = 2\)
Substitute \(d = 2\) in equation (2):
\(2a + 29(2) = 160\)
\(2a + 58 = 160\)
\(2a = 102 \implies a = 51\)
Therefore, the first instalment is Rs. 51.
In simple words: Set up two equations using the sum formula for 40 terms and 30 terms, then solve them simultaneously to find the first instalment.

Exam Tip: Be careful with the fraction of the debt paid versus unpaid. Two thirds of 3600 is 2400, which equals \(S_{30}\).

Question 6. Find the sum of all 3 digit numbers which leave remainder 3 when divided by 5.
Answer: A 3-digit number leaving a remainder of 3 when divided by 5 starts from 103 and ends at 998.
The sequence is \(103, 108, 113, \dots, 998\).
This forms an AP with first term \(a = 103\), common difference \(d = 5\), and last term \(l = 998\).
To find the number of terms \(n\):
\(a_n = a + (n-1)d = 998\)
\(103 + (n-1)5 = 998\)
\((n-1)5 = 998 - 103 = 895\)
\(n - 1 = \frac{895}{5} = 179\)
\(n = 180\)
Now, find the sum of these 180 terms:
\(S_{180} = \frac{n}{2}(a + l)\)
\(S_{180} = \frac{180}{2}(103 + 998)\)
\(S_{180} = 90 \times 1101 = 99090\)
In simple words: Find the first and last 3-digit numbers satisfying the condition, determine total count \(n\), and apply the sum formula.

Exam Tip: Always double-check your first and last numbers by performing division checks (e.g., \(103 \div 5\) gives quotient 20 and remainder 3).

Question 7. Find the value of \(x\) if \(2x + 1\), \(x^2 + x + 1\), \(3x^2 - 3x + 3\) are consecutive terms of an AP.
Answer: Let \(a_1 = 2x + 1\), \(a_2 = x^2 + x + 1\), and \(a_3 = 3x^2 - 3x + 3\).
Since they are consecutive terms of an AP, the common difference is constant:
\(a_2 - a_1 = a_3 - a_2\)
\((x^2 + x + 1) - (2x + 1) = (3x^2 - 3x + 3) - (x^2 + x + 1)\)
Simplify both sides:
\(x^2 + x + 1 - 2x - 1 = 3x^2 - 3x + 3 - x^2 - x - 1\)
\(x^2 - x = 2x^2 - 4x + 2\)
Rearrange all terms to one side:
\(2x^2 - x^2 - 4x + x + 2 = 0\)
\(x^2 - 3x + 2 = 0\)
Factorize the quadratic equation:
\((x - 1)(x - 2) = 0\)
\(x = 1 \text{ or } x = 2\)
In simple words: Use the property that the difference between consecutive terms in an AP is equal to form an equation in terms of \(x\), then solve the quadratic equation.

Question 8. Raghav buys a shop for Rs. 1,20,000. He pays half the balance of the amount in cash and agrees to pay the balance in 12 annual instalments of Rs. 5,000 each. If the rate of interest is 12% and he pays the instalment due for the unpaid amount, find the total cost of the shop.
Answer: Total cost of the shop = Rs. 1,20,000.

Cash paid = Half of the total amount = \(\frac{1,20,000}{2} = \text{Rs. } 60,000\).
Unpaid balance to be paid in 12 annual instalments = Rs. 60,000 (at Rs. 5000 principal per instalment).
- Amount of first instalment = Principal + Interest on full unpaid amount
\(= 5000 + \frac{12}{100} \times 60000 = 5000 + 7200 = 12200\)
- Amount of second instalment = Principal + Interest on remaining unpaid amount (Rs. 55,000)
\(= 5000 + \frac{12}{100} \times 55000 = 5000 + 6600 = 11600\)
- Amount of third instalment = \(5000 + \frac{12}{100} \times 50000 = 5000 + 6000 = 11000\)
The instalments form an AP: \(12200, 11600, 11000, \dots\) up to 12 terms.
Here, first term \(a = 12200\), common difference \(d = 11600 - 12200 = -600\), and \(n = 12\).
Total sum of instalments:
\(S_{12} = \frac{12}{2}[2(12200) + (12-1)(-600)]\)
\(S_{12} = 6[24400 + 11(-600)]\)
\(S_{12} = 6[24400 - 6600]\)
\(S_{12} = 6 \times 17800 = 106800\)
Total cost of the shop = Cash paid + Sum of all instalments
\(= 60000 + 106800 = 166800\)
Therefore, the total cost of the shop is Rs. 1,66,800.
In simple words: Calculate the diminishing interest for each year's unpaid balance, form an AP of the total instalment amounts, sum them up, and add the initial cash payment.

Exam Tip: Note that the principal component of each instalment is fixed at Rs. 5000, so the interest decreases by \(5000 \times 12\% = 600\) every year.

Question 9. Prove that \(a_{m+n} + a_{m-n} = 2a_m\).
Answer: Let the first term of the AP be \(a_1\) and the common difference be \(d\).
The general term is given by \(a_k = a_1 + (k-1)d\).
Write expressions for \(a_{m+n}\), \(a_{m-n}\), and \(a_m\):
\(a_{m+n} = a_1 + (m + n - 1)d \quad \text{--- (1)}\)
\(a_{m-n} = a_1 + (m - n - 1)d \quad \text{--- (2)}\)
\(a_m = a_1 + (m - 1)d\)
Adding equations (1) and (2):
\(a_{m+n} + a_{m-n} = [a_1 + (m + n - 1)d] + [a_1 + (m - n - 1)d]\)
\(= 2a_1 + [m + n - 1 + m - n - 1]d\)
\(= 2a_1 + [2m - 2]d\)
\(= 2a_1 + 2(m - 1)d\)
\(= 2[a_1 + (m - 1)d]\)
\(= 2a_m\)
Hence proved.
In simple words: Expand each term using the standard AP formula, add them together, and factor out 2 to arrive at \(2a_m\).

Exam Tip: This is a standard algebraic proof in AP. Keep your steps clear and align the brackets properly.

Question 10. If the roots of the equation \((b-c)x^2 + (c-a)x + (a-b) = 0\) are equal, show that \(a, b, c\) are in AP.
Answer: For a quadratic equation \(Ax^2 + Bx + C = 0\) to have equal roots, its discriminant must be zero:
\(B^2 - 4AC = 0\)
Here, \(A = b - c\), \(B = c - a\), and \(C = a - b\).
Substitute these into the discriminant condition:
\((c - a)^2 - 4(b - c)(a - b) = 0\)
Expand both terms:
\(c^2 + a^2 - 2ac - 4[ab - b^2 - ac + bc] = 0\)
\(c^2 + a^2 - 2ac - 4ab + 4b^2 + 4ac - 4bc = 0\)
\(a^2 + 4b^2 + c^2 + 2ac - 4ab - 4bc = 0\)
Notice that this expression can be written as a complete square:
\((a - 2b + c)^2 = 0\)
\(a - 2b + c = 0\)
\(a + c = 2b\)
This condition implies that \(b - a = c - b\), which means \(a, b, c\) are in AP.
In simple words: Apply the equal roots condition (\(B^2 = 4AC\)), expand the algebraic expression, factor it into a square, and obtain the standard AP relation \(2b = a + c\).

Exam Tip: Recognizing the expansion of \((a - 2b + c)^2\) saves significant time compared to other reduction methods.

Question 11. Balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second two balls and so on. If 669 more balls are added, then all the balls can be arranged in the shape of a square and each of its sides then contains 8 balls less than each side of the triangle. Find the initial number of balls.
Answer: Let \(n\) be the number of balls in each side of the triangular arrangement.
The total number of balls in the triangular arrangement is the sum of natural numbers from 1 to \(n\):
\(\text{Total balls in triangle} = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}\)
The number of balls in each side of the square is \(n - 8\).
The total number of balls in the square arrangement is:
\(\text{Total balls in square} = (n - 8)^2\)
Given that 669 more balls are added to the triangle to form the square:
\(\frac{n(n+1)}{2} + 669 = (n - 8)^2\)
Multiply the entire equation by 2 to clear the fraction:
\(n(n+1) + 1338 = 2(n^2 - 16n + 64)\)
\(n^2 + n + 1338 = 2n^2 - 32n + 128\)
Rearrange all terms to one side:
\(n^2 - 33n - 1210 = 0\)
Factorize the quadratic equation:
\((n - 55)(n + 22) = 0\)
\(n = 55 \text{ or } n = -22\)
Since the number of balls cannot be negative, we take \(n = 55\).
Initial number of balls in the triangle = \(\frac{55(55+1)}{2} = \frac{55 \times 56}{2} = 55 \times 28 = 1540\).
In simple words: Set up an equation equating the triangular sum plus 669 to the square of the side length, solve the resulting quadratic equation, and calculate the total balls.

Exam Tip: Always reject negative roots in physical word problems involving counts and quantities.

Question 12. Find the sum of \(\left(1 - \frac{1}{n}\right) + \left(1 - \frac{2}{n}\right) + \left(1 - \frac{3}{n}\right) + \dots\) upto \(n\) terms.
Answer: The given series is:
\(\left(1 - \frac{1}{n}\right) + \left(1 - \frac{2}{n}\right) + \left(1 - \frac{3}{n}\right) + \dots\) up to \(n\) terms.
Separate the terms into two distinct groups:
\(= (1 + 1 + 1 + \dots \text{ upto } n \text{ terms}) - \left(\frac{1}{n} + \frac{2}{n} + \frac{3}{n} + \dots \text{ upto } n \text{ terms}\right)\)
For the first group, the sum of \(n\) ones is \(n\).
For the second group, factor out \(\frac{1}{n}\):
\(= n - \frac{1}{n}(1 + 2 + 3 + \dots + n)\)
Using the formula for the sum of first \(n$ natural numbers, \(1 + 2 + \dots + n = \frac{n(n+1)}{2}\):
\(= n - \frac{1}{n} \left[ \frac{n(n+1)}{2} \right]\)
Simplify the expression:
\(= n - \frac{n+1}{2}\)
\(= \frac{2n - (n+1)}{2}\)
\(= \frac{2n - n - 1}{2}\)
\(= \frac{n - 1}{2}\)
In simple words: Group the constant terms together and factor out the fractional terms into a standard arithmetic progression sum.

Exam Tip: Splitting mixed series into simpler standard series is a powerful technique for evaluation.

Question 13. If the following terms form an AP, find the common difference and write the next 3 terms: \(3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}, \dots\)
Answer: Given AP: \(3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}, \dots\)
The common difference \(d\) is calculated as:
\(d = (3 + \sqrt{2}) - 3 = \sqrt{2}\)
To find the next three terms, add the common difference \(\sqrt{2}\) successively:
- Next term 1: \((3 + 3\sqrt{2}) + \sqrt{2} = 3 + 4\sqrt{2}\)
- Next term 2: \((3 + 4\sqrt{2}) + \sqrt{2} = 3 + 5\sqrt{2}\)
- Next term 3: \((3 + 5\sqrt{2}) + \sqrt{2} = 3 + 6\sqrt{2}\)
Therefore, the common difference is \(\sqrt{2}\) and the next three terms are \(3 + 4\sqrt{2}\), \(3 + 5\sqrt{2}\), and \(3 + 6\sqrt{2}\).
In simple words: Subtract the first term from the second term to find the common difference, then add it repeatedly to get the subsequent terms.

Exam Tip: Surds behave like algebraic variables; treat \(\sqrt{2}\) like \(x\) when combining like terms.

Question 14. Find the sum of \(a+b, a-b, a-3b, \dots\) to 22 terms.
Answer: Given AP: \(a+b, a-b, a-3b, \dots\)
First term \(A = a + b\)
Common difference \(d = (a - b) - (a + b) = a - b - a - b = -2b\)
Number of terms \(n = 22\)
Using the sum formula \(S_n = \frac{n}{2}[2A + (n-1)d]\):
\(S_{22} = \frac{22}{2}[2(a + b) + (22-1)(-2b)]\)
\(S_{22} = 11[2a + 2b + 21(-2b)]\)
\(S_{22} = 11[2a + 2b - 42b]\)
\(S_{22} = 11[2a - 40b]\)
\(S_{22} = 22a - 440b\)
In simple words: Identify the first term and common difference carefully, taking note of negative signs, and substitute into the sum formula.

RAnswer: Simplify each radical expression by factoring out perfect squares:
\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\)
\(\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}\)
\(\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}\)
\(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\)
The sequence in simplified form is \(2\sqrt{3}, 3\sqrt{3}, 4\sqrt{3}, 5\sqrt{3}, \dots\)
This is an AP with common difference \(d = \sqrt{3}\).
The next two terms are:
- Fifth term: \(6\sqrt{3} = \sqrt{36 \times 3} = \sqrt{108}\)
- Sixth term: \(7\sqrt{3} = \sqrt{49 \times 3} = \sqrt{147}\)
In simple words: Simplify the square roots to reveal the underlying arithmetic progression pattern.

Exam Tip: Always simplify surds first before attempting to find the common difference.

Question 16. If the \(p^{\text{th}}\) term of an AP is \(q\) and the \(q^{\text{th}}\) term is \(p\), prove that its \(n^{\text{th}}\) term is \((p + q - n)\).
Answer: Let the first term of the AP be \(a\) and the common difference be \(d\).
According to the given conditions:
\(a_p = a + (p-1)d = q \quad \text{--- (1)}\)
\(a_q = a + (q-1)d = p \quad \text{--- (2)}\)
Subtract equation (2) from equation (1):
\((p - q)d = q - p\)
\(d(p - q) = -(p - q)\)
\(d = -1\)
Substitute \(d = -1\) into equation (1):
\(a + (p - 1)(-1) = q\)
\(a - p + 1 = q \implies a = q + p - 1\)
Now, find the \(n^{\text{th}}\) term (\(a_n\)):
\(a_n = a + (n - 1)d\)
Substitute \(a = q + p - 1\) and \(d = -1\):
\(a_n = (q + p - 1) + (n - 1)(-1)\)
\(a_n = q + p - 1 - n + 1\)
\(a_n = p + q - n\)
Hence proved.
In simple words: Set up simultaneous equations from the given terms, solve for \(a\) and \(d\), and substitute them into the \(n^{\text{th}}\) term formula.

Exam Tip: This is a classic standard result. Memorize the technique of subtracting the two equations to quickly find \(d = -1\).

Question 17. If \(\frac{1}{x+2}, \frac{1}{x+3}, \frac{1}{x+5}\) are in AP, find \(x\).
Answer: Since the terms are in AP, the difference between consecutive terms is equal:
\(\frac{1}{x+3} - \frac{1}{x+2} = \frac{1}{x+5} - \frac{1}{x+3}\)
Take the LCM on both sides:
\(\frac{(x+2) - (x+3)}{(x+3)(x+2)} = \frac{(x+3) - (x+5)}{(x+5)(x+3)}\)
Simplify the numerators:
\(\frac{-1}{(x+3)(x+2)} = \frac{-2}{(x+5)(x+3)}\)
Cancel \((x+3)\) from both denominators and cross-multiply:
\(-1(x + 5) = -2(x + 2)\)
\(-x - 5 = -2x - 4\)
\(2x - x = 5 - 4\)
\(x = 1\)
In simple words: Use the AP condition of equal common differences, simplify the fractions, and solve the linear equation for \(x\).

Exam Tip: You can cancel common factors across denominators only if they are non-zero.

Question 18. Find the middle term of the AP \(1, 8, 15, \dots, 505\).
Answer: Given AP: \(1, 8, 15, \dots, 505\)
Here, first term \(a = 1\), common difference \(d = 8 - 1 = 7\), and last term \(l = 505\).
First, find the total number of terms \(n\):
\(a_n = a + (n-1)d = 505\)
\(1 + (n-1)7 = 505\)
\((n-1)7 = 504\)
\(n - 1 = \frac{504}{7} = 72\)
\(n = 73\)
Since \(n = 73\) is an odd number, there is a single middle term, which is the \(\left(\frac{73+1}{2}\right)^{\text{th}}\) term, i.e., the \(37^{\text{th}}\) term.
Calculate \(a_{37}\):
\(a_{37} = a + 36d\)
\(a_{37} = 1 + 36(7)\)
\(a_{37} = 1 + 252 = 253\)
Therefore, the middle term is 253.
In simple words: Find the total number of terms first. If \(n\) is odd, use the middle term formula \(\frac{n+1}{2}\) to find its position.

Exam Tip: If \(n\) is even, there will be two middle terms given by \(\frac{n}{2}\) and \(\frac{n}{2} + 1\).

Question 19. Find the common difference of an AP whose first term is 100 and sum of whose first 6 terms is 5 times the sum of next 6 terms.
Answer: Given that the first term \(a = 100\).
The sum of the first 6 terms is equal to 5 times the sum of the next 6 terms:
\(a_1 + a_2 + \dots + a_6 = 5(a_7 + a_8 + \dots + a_{12})\)
Using the sum formula for 6 terms: sum of first 6 terms is \(S_6\), and the sum of the next 6 terms is \(S_{12} - S_6\).
\(S_6 = 5(S_{12} - S_6)\)
\(6S_6 = 5S_{12}\)
Substitute the sum formulas:\(S_6 = \frac{6}{2}[2a + 5d]\) and \(S_{12} = \frac{12}{2}[2a + 11d]\):
\(6 \times \frac{6}{2}[2a + 5d] = 5 \times \frac{12}{2}[2a + 11d]\)
\(18[2a + 5d] = 30[2a + 11d]\)
Divide both sides by 6:
\(3[2a + 5d] = 5[2a + 11d]\)
\(6a + 15d = 10a + 55d\)
Rearrange terms:
\(6a - 10a = 55d - 15d\)
\(-4a = 40d \implies a = -10d\)
Given that \(a = 100\):
\(100 = -10d \implies d = -10\)
Therefore, the common difference is \(-10\).
In simple words: Translate the condition into terms of \(S_6\) and \(S_{12}\), simplify the equation, and substitute \(a = 100\) to find \(d\).

Exam Tip: Recognizing that the next 6 terms equals \(S_{12} - S_6\) drastically simplifies the setup.

Question 20. Find the sum of all natural numbers between 101 and 304 which are divisible by 3 or 5.
Answer: We need to find the sum of numbers between 101 and 304 divisible by 3 or 5.
- Numbers divisible by 3: \(102, 105, \dots, 303\)
Here, \(a = 102\), \(d = 3\), \(l = 303\).
\(303 = 102 + (n-1)3 \implies 201 = (n-1)3 \implies n - 1 = 67 \implies n = 68\)
Sum \(S_3 = \frac{68}{2}(102 + 303) = 34 \times 405 = 13770\)
- Numbers divisible by 5: \(105, 110, \dots, 300\)
Here, \(a = 105\), \(d = 5\), \(l = 300\).
\(300 = 105 + (n-1)5 \implies 195 = (n-1)5 \implies n - 1 = 39 \implies n = 40\)
Sum \(S_5 = \frac{40}{2}(105 + 300) = 20 \times 405 = 8100\)
- Numbers divisible by both 3 and 5 (i.e., divisible by 15): \(105, 120, \dots, 300\)
Here, \(a = 105\), \(d = 15\), \(l = 300\).
\(300 = 105 + (n-1)15 \implies 195 = (n-1)15 \implies n - 1 = 13 \implies n = 14\)
Sum \(S_{15} = \frac{14}{2}(105 + 300) = 7 \times 405 = 2835\)
Using the Principle of Inclusion-Exclusion for sums:
\(\text{Total Sum} = S_3 + S_5 - S_{15}\)
\(\text{Total Sum} = 13770 + 8100 - 2835 = 19035\)
In simple words: Find the sum of multiples of 3, add the sum of multiples of 5, and subtract the sum of common multiples of 15 to avoid double counting.

Exam Tip: Always subtract the intersection set (LCM multiples) when calculating unions of divisibility.

Question 21. The ratio of the sum of first \(n\) terms of two APs is \((7n + 1):(4n + 27)\). Find the ratio of their \(11^{\text{th}}\) terms.
Answer: Let the first terms of the two APs be \(a_1, a_2\) and their common differences be \(d_1, d_2\) respectively.
Given the ratio of the sum of \(n\) terms:
\(\frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{7n + 1}{4n + 27}\)
\(\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{7n + 1}{4n + 27} \quad \text{--- (1)}\)
We need to find the ratio of their \(11^{\text{th}}\) terms, which is given by \(\frac{a_1 + 10d_1}{a_2 + 10d_2}\).
To match the coefficient of \(d_1\) and \(d_2\) to 10, we set \(\frac{n-1}{2} = 10\), which gives \(n - 1 = 20 \implies n = 21\).
Substitute \(n = 21\) into equation (1):
\(\frac{2a_1 + 20d_1}{2a_2 + 20d_2} = \frac{7(21) + 1}{4(21) + 27}\)
\(\frac{2(a_1 + 10d_1)}{2(a_2 + 10d_2)} = \frac{147 + 1}{84 + 27}\)
\(\frac{a_1 + 10d_1}{a_2 + 10d_2} = \frac{148}{111}\)
Divide numerator and denominator by 37:
\(\frac{a_1 + 10d_1}{a_2 + 10d_2} = \frac{4}{3}\)
Therefore, the ratio of their \(11^{\text{th}}\) terms is \(4:3\).
In simple words: Substitute \(n = 2q - 1\) where \(q\) is the desired term index (here \(q = 11\), so \(n = 21\)) into the ratio of sums to directly get the ratio of the \(q^{\text{th}}\) terms.

Exam Tip: This is a standard shortcut: to find the ratio of the \(n^{\text{th}}\) term, substitute \(2n - 1\) in place of \(n\) in the ratio of sums.

Question 22. If there are \((2n + 1)\) terms in an AP, prove that the ratio of the sum of odd terms and the sum of even terms is \((n + 1) : n\).
Answer: Let the AP have first term \(a\) and common difference \(d\). Total terms = \(2n + 1\).
- Odd terms are \(a_1, a_3, a_5, \dots, a_{2n+1}\).
These form an AP where the first term is \(a_1 = a\), the common difference is \(2d$, and the number of terms is \(n + 1\).
Sum of odd terms (\(S_{\text{odd}}\)):
\(S_{\text{odd}} = \frac{n+1}{2}[2a + (n+1-1)(2d)]\)
\(S_{\text{odd}} = \frac{n+1}{2}[2a + 2nd] = (n+1)(a + nd)\)
- Even terms are \(a_2, a_4, a_6, \dots, a_{2n}\).
These form an AP where the first term is \(a_2 = a + d\), the common difference is \(2d\), and the number of terms is \(n\).
Sum of even terms (\(S_{\text{even}}\)):
\(S_{\text{even}} = \frac{n}{2}[2(a + d) + (n-1)(2d)]\)
\(S_{\text{even}} = \frac{n}{2}[2a + 2d + 2nd - 2d] = \frac{n}{2}[2a + 2nd] = n(a + nd)\)
Taking the ratio of the sum of odd terms to the sum of even terms:
\(\frac{S_{\text{odd}}}{S_{\text{even}}} = \frac{(n+1)(a + nd)}{n(a + nd)} = \frac{n+1}{n}\)
Hence proved.
In simple words: Separate the series into odd-positioned and even-positioned terms, find the number of terms and common difference for each new sub-series, and compute their ratio.

Exam Tip: Note that the odd terms sub-series has \(n+1\) terms while the even terms sub-series has \(n\) terms.

Question 23. Find the sum of all natural numbers amongst first one thousand numbers which are neither divisible by 2 nor by 5.
Answer: We need to find the sum of natural numbers from 1 to 1000 that are not divisible by 2 or 5.
- Sum of all integers from 1 to 1000:
\(S_{\text{total}} = \frac{1000}{2}(1 + 1000) = 500 \times 1001 = 500500$
- Sum of integers divisible by 2 (even integers: \(2, 4, 6, \dots, 1000\)):
Number of terms \(n = 500\).
\(S_2 = \frac{500}{2}(2 + 1000) = 250 \times 1002 = 250500\)
- Sum of integers divisible by 5 (\(5, 10, 15, \dots, 1000\)):
Number of terms \(n = \frac{1000}{5} = 200\).
\(S_5 = \frac{200}{2}(5 + 1000) = 100 \times 1005 = 100500\)
- Sum of integers divisible by both 2 and 5 (i.e., divisible by 10: \(10, 20, 30, \dots, 1000\)):
Number of terms \(n = \frac{1000}{10} = 100\).
\(S_{10} = \frac{100}{2}(10 + 1000) = 50 \times 1010 = 50500\)
Using the Principle of Inclusion-Exclusion, the sum of numbers divisible by 2 or 5 is:
\(S_{2 \text{ or } 5} = S_2 + S_5 - S_{10}\)
\(S_{2 \text{ or } 5} = 250500 + 100500 - 50500 = 300500\)
Finally, subtract this from the total sum to find numbers neither divisible by 2 nor by 5:
\(\text{Required Sum} = S_{\text{total}} - S_{2 \text{ or } 5}\)
\(\text{Required Sum} = 500500 - 300500 = 200000\)
In simple words: Subtract the sum of numbers divisible by 2 or 5 (accounting for overlap by adding back multiples of 10) from the total sum of the first 1000 numbers.

Exam Tip: Be careful with the union and intersection formulas when filtering out multiple divisors.