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Question 1 / 20
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1
Calculate the sum of the series: (1 - 1/n) + (1 - 2/n) + (1 - 3/n) + ... up to n terms.
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Solution: Step 1: Identify the pattern of the series. Each term is of the form (1 - k/n), where k ranges from 1 to n. Step 2: Rewrite the sum by separating the integer '1' part and the fractional part: Sum = (1 + 1 + ... + 1 (n times)) - (1/n + 2/n + 3/n + ... + n/n). Step 3: The sum of 'n' ones is simply n. Step 4: Factor out 1/n from the fractional sum: Fractional sum = (1/n) * (1 + 2 + 3 + ... + n). Step 5: Apply the formula for the sum of the first 'n' natural numbers, which is Sum_k = k(k+1)/2. For 'n' terms, the sum is n(n+1)/2. Fractional sum = (1/n) * [n(n+1)/2]. Step 6: Simplify the fractional sum: n cancels out, leaving (n+1)/2. Step 7: Substitute these results back into the overall sum equation: Sum = n - (n+1)/2. Step 8: Combine the terms by finding a common denominator: Sum = (2n)/2 - (n+1)/2 = (2n - (n+1))/2. Step 9: Simplify the numerator: (2n - n - 1)/2 = (n - 1)/2. Step 10: Therefore, the sum of the series is (n - 1)/2.
2
In an arithmetic series, the sum of its first 10 terms is 390, and its third term is 19. What is the first term of this series?
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Solution: Step 1: Write down the formula for the nth term of an AP. a_n = a + (n - 1)d, where 'a' is the first term and 'd' is the common difference. Step 2: Use the given information about the third term. a_3 = a + (3 - 1)d = a + 2d. Given a_3 = 19, so: a + 2d = 19 (Equation 1). Step 3: Write down the formula for the sum of n terms of an AP. S_n = n/2 × [2a + (n - 1)d]. Step 4: Use the given information about the sum of 10 terms. S_10 = 10/2 × [2a + (10 - 1)d] = 5 × [2a + 9d]. Given S_10 = 390, so: 5(2a + 9d) = 390. Divide by 5: 2a + 9d = 78 (Equation 2). Step 5: Solve the system of linear equations (Equation 1 and Equation 2). From Equation 1, express 'a' in terms of 'd': a = 19 - 2d. Substitute this into Equation 2: 2(19 - 2d) + 9d = 78 38 - 4d + 9d = 78 38 + 5d = 78 5d = 78 - 38 5d = 40 d = 8. Step 6: Substitute the value of d back into Equation 1 to find 'a'. a + 2(8) = 19 a + 16 = 19 a = 19 - 16 a = 3. The first term of the series is 3.
3
Calculate the sum of all even numbers located between 1 and 31.
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Solution: Step 1: Identify the even numbers between 1 and 31. These are 2, 4, 6, ..., 30. Step 2: Recognize this as an Arithmetic Progression (AP) with the first term (a) = 2, common difference (d) = 2, and last term (l) = 30. Step 3: Find the number of terms (n) using the formula l = a + (n-1)d. 30 = 2 + (n-1)2 28 = (n-1)2 14 = n-1 n = 15. Step 4: Calculate the sum of the AP using the formula Sn = n/2 * (a + l). Sn = 15/2 * (2 + 30) Sn = 15/2 * 32 Sn = 15 * 16 Sn = 240. Step 5: The sum of even numbers between 1 and 31 is 240.
4
Calculate the sum of the initial 13 terms of an arithmetic progression, given that its first term is -10 and its last term is 26.
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Solution: Step 1: Identify the given values: first term (a) = -10, last term (l) = 26, number of terms (n) = 13. Step 2: Use the formula for the sum of an arithmetic progression: S_n = n/2 * (a + l). Step 3: Substitute the values into the formula: S_13 = 13/2 * (-10 + 26). Step 4: Simplify the expression: S_13 = 13/2 * (16). Step 5: Calculate the sum: S_13 = 13 * 8 = 104.
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What is the common difference of the Arithmetic Progression: 1/3, (1-3b)/3, (1-6b)/3, ...?
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Solution: Step 1: Identify the first term (t_1) and the second term (t_2) of the A.P.: * t_1 = 1 / 3 * t_2 = (1 - 3b) / 3 Step 2: The common difference (d) of an A.P. is found by subtracting any term from its succeeding term. So, d = t_2 - t_1. * d = (1 - 3b) / 3 - 1 / 3 Step 3: Since the terms have a common denominator, combine the numerators: * d = (1 - 3b - 1) / 3 * d = -3b / 3 Step 4: Simplify the expression by cancelling '3': * d = -b
6
Find the sum of the series: (1) + (1 + 1) + (1 + 1 + 1) + ... + (sum of 1, (n-1) times).
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Solution: Step 1: Analyze the terms of the series. The first term is 1. The second term is (1 + 1) = 2. The third term is (1 + 1 + 1) = 3. ... The last term is (1 + 1 + ... + 1 (n-1 times)) = (n-1). Step 2: This series is the sum of the first (n-1) natural numbers: 1 + 2 + 3 + ... + (n-1). Step 3: Use the formula for the sum of the first 'k' natural numbers: S_k = k(k+1)/2. Step 4: In this case, k = (n-1). Substitute this into the formula: Sum = (n-1)((n-1)+1)/2 Sum = (n-1)n/2 Sum = n(n-1)/2.
7
How many distinct ways can the letters of the word 'DETAIL' be organized so that vowels are exclusively placed in odd-numbered positions?
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Solution: Step 1: Identify total positions and letter types. The word 'DETAIL' has 6 letters. Vowels: E, A, I (3 vowels). Consonants: D, T, L (3 consonants). Step 2: Identify odd positions. The odd positions are 1st, 3rd, 5th. There are 3 odd positions. Step 3: Arrange vowels. The 3 vowels (E, A, I) must occupy the 3 odd positions (1, 3, 5). The number of ways to arrange 3 distinct items in 3 distinct positions is 3P3 = 3! = 3 × 2 × 1 = 6 ways. Step 4: Arrange consonants. The remaining 3 consonants (D, T, L) must occupy the remaining 3 even positions (2, 4, 6). The number of ways to arrange 3 distinct items in 3 distinct positions is 3P3 = 3! = 3 × 2 × 1 = 6 ways. Step 5: Calculate total arrangements. To find the total number of ways, multiply the number of ways to arrange vowels by the number of ways to arrange consonants: 6 × 6 = 36 ways.
8
Identify the next term in the Arithmetic Progression (A.P.): √7, √28, √63, ...
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Solution: Step 1: Simplify the given terms of the A.P. √7 = √7 √28 = √(4 * 7) = 2√7 √63 = √(9 * 7) = 3√7 Step 2: Observe the pattern and identify the common difference (d). The sequence in its simplified form is √7, 2√7, 3√7, ... This is an A.P. where the first term (a) = √7. The common difference (d) = 2√7 - √7 = √7. Step 3: Calculate the next term in the sequence. The given terms are the 1st, 2nd, and 3rd terms. We need to find the 4th term. 4th term = 3rd term + d 4th term = 3√7 + √7 = 4√7 Step 4: Convert the next term back to the original format (square root). 4√7 = √(4^2 * 7) = √(16 * 7) = √112
9
Find the 15th term of the arithmetic progression (A.P.) defined by the sequence x - 7, x - 2, x + 3, and so on.
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Solution: Step 1: Identify the first term (a) and calculate the common difference (d). a = x - 7 d = (x - 2) - (x - 7) = x - 2 - x + 7 = 5 Step 2: Use the formula for the nth term of an A.P.: a_n = a + (n-1)d. Step 3: Substitute the values for 'a', 'd', and n=15: a_15 = (x - 7) + (15 - 1) * 5 a_15 = (x - 7) + 14 * 5 a_15 = x - 7 + 70 a_15 = x + 63.
10
A single bacterium divides into eight bacteria for the subsequent generation. However, only 50% of any given generation successfully produce the next. If the seventh generation consists of 4096 million bacteria, what was the initial number of bacteria in the first generation?
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Solution: Step 1: Determine the multiplication factor from one generation to the next. Each bacterium splits into 8, but only 50% survive to produce the next generation. So, the effective multiplication factor per generation is 50% of 8 = 0.5 * 8 = 4. Step 2: This represents a geometric progression where each term is 4 times the previous term. Step 3: Let the number of bacteria in the first generation be 'x'. The number of bacteria in the n-th generation (a_n) is given by a_n = x * (factor)^(n-1). Step 4: For the 7th generation, a_7 = x * 4^(7-1) = x * 4^6. Step 5: We are given a_7 = 4096 million. So, x * 4^6 = 4096. Step 6: Calculate 4^6: 4^6 = (4^3)^2 = 64^2 = 4096. Step 7: Substitute this back into the equation: x * 4096 = 4096. Step 8: Solve for x: x = 4096 / 4096 = 1 million.
11
A question paper has three sections containing 4, 5, and 6 questions, respectively. It is mandatory to attempt at least one question from each section, but a candidate is not required to attempt all questions. Determine the total number of distinct ways a candidate can attempt the questions.
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Solution: Step 1: For the first section with 4 questions, each question can either be attempted or left unattempted. This provides 2 choices per question, totaling 2^4 ways. Step 2: Since at least one question must be attempted from this section, subtract the case where no questions are attempted (1 way). So, ways for Section 1 = 2^4 - 1 = 16 - 1 = 15. Step 3: Similarly, for the second section with 5 questions, the number of ways is 2^5 - 1 = 32 - 1 = 31. Step 4: For the third section with 6 questions, the number of ways is 2^6 - 1 = 64 - 1 = 63. Step 5: Since the selections for each section are independent, multiply the number of ways for each section to find the total number of ways. Step 6: Total ways = (2^4 - 1) * (2^5 - 1) * (2^6 - 1) = 15 * 31 * 63. Step 7: Calculate the product: 15 * 31 = 465. Step 8: Multiply by 63: 465 * 63 = 29295.
12
At a gathering, every person present shakes hands with every other person exactly once. If a total of 105 handshakes occurred, how many people were in attendance at the party?
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Solution: Step 1: Let 'x' be the total number of persons present at the party. Step 2: A handshake involves two people. Since the order of people in a handshake does not matter, this is a combination problem. Step 3: The number of handshakes is given by the combination formula C(x, 2) = x × (x - 1) / 2. Step 4: Set up the equation using the given number of handshakes: x × (x - 1) / 2 = 105. Step 5: Multiply both sides by 2: x × (x - 1) = 210. Step 6: Expand and rearrange into a quadratic equation: x^2 - x - 210 = 0. Step 7: Factorize the quadratic equation: (x - 15)(x + 14) = 0. Step 8: This yields two possible values for x: x = 15 or x = -14. Step 9: Since the number of persons cannot be negative, discard x = -14. Step 10: Therefore, 15 persons were present at the party.
13
Six fair coins are tossed concurrently. How many of the possible outcomes will result in at most three coins showing heads?
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Solution: Step 1: We are tossing 6 fair coins. Each toss can result in a Head (H) or a Tail (T). Step 2: The condition "at most three of the coins turn up as heads" means we need to count the outcomes where the number of heads is 0, 1, 2, or 3. Step 3: This is a combination problem for each case, as the order of individual coin outcomes for a given number of heads does not matter. Step 4: Calculate the number of ways for each case: Case 1: 0 Heads (all 6 are tails). Number of ways = C(6, 0) = 1. Case 2: 1 Head (and 5 tails). Number of ways = C(6, 1) = 6. Case 3: 2 Heads (and 4 tails). Number of ways = C(6, 2) = (6 × 5) / (2 × 1) = 15. Case 4: 3 Heads (and 3 tails). Number of ways = C(6, 3) = (6 × 5 × 4) / (3 × 2 × 1) = 20. Step 5: Since these cases are mutually exclusive, add the number of ways for each case to find the total number of outcomes. Step 6: Total outcomes = 1 (for 0 heads) + 6 (for 1 head) + 15 (for 2 heads) + 20 (for 3 heads) = 42. Step 7: Therefore, there are 42 outcomes where at most three of the coins turn up as heads.
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How many terms of the Arithmetic Progression 3, 7, 11, 15, ... must be summed to achieve a total of 406?
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Solution: Step 1: Identify the given values from the A.P.: * First term (a) = 3 * Common difference (d) = 7 - 3 = 4 * Sum of 'n' terms (S_n) = 406 Step 2: Use the formula for the sum of 'n' terms of an A.P.: * S_n = n/2 * [2a + (n-1)d] Step 3: Substitute the known values into the formula: * 406 = n/2 * [2(3) + (n-1)4] * 406 = n/2 * [6 + 4n - 4] * 406 = n/2 * [4n + 2] Step 4: Simplify the equation and form a quadratic equation: * 406 = n(2n + 1) * 406 = 2n² + n * 2n² + n - 406 = 0 Step 5: Solve the quadratic equation for 'n' using factorization: * Find two numbers that multiply to 2 * (-406) = -812 and add up to 1 (the coefficient of n). These numbers are 29 and -28. * 2n² + 29n - 28n - 406 = 0 * n(2n + 29) - 14(2n + 29) = 0 * (2n + 29)(n - 14) = 0 Step 6: Determine the valid value for 'n': * n - 14 = 0 ⇒ n = 14 * 2n + 29 = 0 ⇒ n = -29/2 * Since the number of terms cannot be negative or a fraction, n = 14 is the correct answer.
15
If the sum of squares from 1 to 10 (1² + 2² + ... + 10²) is 385, what is the value of the sum of squares of even numbers from 2 to 20 (2² + 4² + ... + 20²)?
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Solution: Step 1: The given sum is S₁ = (1² + 2² + 3² + ... + 10²) = 385. Step 2: The sum to be found is S₂ = (2² + 4² + 6² + ... + 20²). Step 3: Each term in S₂ can be written as (2n)² = 4n². Step 4: Rewrite S₂ using this property: S₂ = (2×1)² + (2×2)² + (2×3)² + ... + (2×10)². Step 5: Factor out 2² (which is 4) from each term: S₂ = 2² × (1² + 2² + 3² + ... + 10²). Step 6: Substitute the value of S₁: S₂ = 4 × S₁ = 4 × 385. Step 7: Calculate the final value: S₂ = 1540.
16
Determine the 15th term of the sequence: 20, 15, 10, ...
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Solution: Step 1: Identify the first term (a) and common difference (d) of the arithmetic sequence. a = 20 d = 15 - 20 = -5 Step 2: Use the formula for the nth term of an AP: T_n = a + (n - 1)d. For the 15th term (n=15): T15 = 20 + (15 - 1) * (-5) Step 3: Calculate the value: T15 = 20 + (14) * (-5) T15 = 20 - 70 T15 = -50 Step 4: The 15th term of the sequence is -50.
17
Calculate the sum of the arithmetic series: -64, -66, -68, ..., up to -100.
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Solution: Step 1: Identify the parameters of the arithmetic progression (A.P.). First term (a) = -64 Common difference (d) = -66 - (-64) = -2 Last term (a_n) = -100 Step 2: Find the number of terms (n) using the formula a_n = a + (n-1)d. -100 = -64 + (n-1)(-2) -100 + 64 = -2(n-1) -36 = -2(n-1) 18 = n-1 n = 19 Step 3: Calculate the sum of the series using the formula S_n = n/2 * (a + a_n). S_19 = 19/2 * (-64 + (-100)) S_19 = 19/2 * (-164) S_19 = 19 * (-82) S_19 = -1558.
18
Four numbers in an Arithmetic Progression sum to 50. If the largest of these numbers is four times the smallest, what are the four numbers?
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Solution: Step 1: Let the four numbers in the A.P. be represented as a - 3d, a - d, a + d, and a + 3d. This representation simplifies calculations, especially when dealing with sums and properties involving extremes. Step 2: Use the first given condition: The sum of the four numbers is 50. * (a - 3d) + (a - d) + (a + d) + (a + 3d) = 50 * Combine like terms: 4a = 50 * Solve for 'a': a = 50 / 4 = 12.5 Step 3: Use the second given condition: The greatest number is 4 times the least number. * Greatest number = a + 3d * Least number = a - 3d * a + 3d = 4 * (a - 3d) Step 4: Solve the equation from Step 3 for 'd', substituting the value of 'a' found in Step 2: * a + 3d = 4a - 12d * 3d + 12d = 4a - a * 15d = 3a * Divide by 3: 5d = a * Substitute a = 12.5: 5d = 12.5 * Solve for 'd': d = 12.5 / 5 = 2.5 Step 5: Calculate the four numbers using a = 12.5 and d = 2.5: * First number: a - 3d = 12.5 - 3(2.5) = 12.5 - 7.5 = 5 * Second number: a - d = 12.5 - 2.5 = 10 * Third number: a + d = 12.5 + 2.5 = 15 * Fourth number: a + 3d = 12.5 + 3(2.5) = 12.5 + 7.5 = 20 Step 6: The four numbers are 5, 10, 15, 20.
19
An arithmetic progression's 3rd term is -13 and its 8th term is 2. What is the 14th term?
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Solution: Step 1: Formulate equations from the given terms using the AP formula: T_n = a + (n-1)d. For the 3rd term: a + 2d = -13 (Equation 1) For the 8th term: a + 7d = 2 (Equation 2) Step 2: Subtract Equation 1 from Equation 2 to find the common difference (d): (a + 7d) - (a + 2d) = 2 - (-13) 5d = 15 d = 3 Step 3: Substitute d = 3 into Equation 1 to find the first term (a): a + 2(3) = -13 a + 6 = -13 a = -19 Step 4: Calculate the 14th term (T14) using the formula T_n = a + (n-1)d: T14 = -19 + (14-1) * 3 T14 = -19 + 13 * 3 T14 = -19 + 39 T14 = 20 Step 5: The 14th term of the arithmetic progression is 20.
20
Given that the average of 'n' numbers is 'a'. If the first number is increased by 2, the second by 4, the third by 8, and so on (each increase being twice the previous one), what is the average of these modified 'n' numbers?
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Solution: Step 1: The initial sum of 'n' numbers is na (since average = a). Step 2: The increases applied to the numbers are 2, 4, 8, ..., which form a geometric progression (GP) with first term A=2 and common ratio R=2. Step 3: Calculate the sum of this GP for 'n' terms using the formula S_n = A * (R^n - 1) / (R - 1). Step 4: Substitute the values: S_n = 2 * (2^n - 1) / (2 - 1) = 2(2^n - 1). Step 5: The new total sum of the numbers is the original sum plus the sum of increases: na + 2(2^n - 1). Step 6: The average of the new numbers is the new total sum divided by 'n': [na + 2(2^n - 1)] / n. Step 7: Simplify the expression: a + [2(2^n - 1)] / n.
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