This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions for problems 10, 11, and 12. Problem 10: Evaluate the following sums: _k=1^20 (3k+2). Step 1: Use the linearity property of summation to split the sum. _k=1^20 (3k+2) = _k=1^20 3k + _k=1^20 2 Step 2: Factor out the constant from the first sum and apply the sum formulas. Recall the formulas: _k=1^n k = (n(n+1))/(2) and _k=1^n c = nc. Here, n=20. 3 _k=1^20 k + _k=1^20 2 = 3 ((20(20+1))/(2)) + (2 × 20) Step 3: Calculate the values. = 3 ((20 × 21)/(2)) + 40 = 3 (10 × 21) + 40 = 3(210) + 40 = 630 + 40 = 670 The final answer is 670. Problem 11: Evaluate _k=1^20 (k^2+k). Step 1: Use the linearity property of summation to split the sum. _k=1^20 (k^2+k) = _k=1^20 k^2 + _k=1^20 k Step 2: Apply the sum formulas. Recall the formulas: _k=1^n k^2 = (n(n+1)(2n+1))/(6) and _k=1^n k = (n(n+1))/(2). Here, n=20. For _k=1^20 k^2: = (20(20+1)(2 × 20 + 1))/(6) = (20(21)(41))/(6) = (420 × 41)/(6) = (17220)/(6) = 2870 For _k=1^20 k: = (20(20+1))/(2) = (20 × 21)/(2) = 10 × 21 = 210 Step 3: Add the results. = 2870 + 210 = 3080 The final answer is 3080. Problem 12: The 13th and 16th terms of an A.P. are 80 and 98 respectively. Find the first term, the common difference and the last term if there are 20 terms. Step 1: Set up equations using the formula for the n-th term of an A.P., a_n = a + (n-1)d. Given a_13 = 80: a + (13-1)d = 80 a + 12d = 80 (1) Given a_16 = 98: a + (16-1)d = 98 a + 15d = 98 (2) Step 2: Solve the system of equations to find the common difference d. Subtract equation (1) from equation (2): (a + 15d) - (a + 12d) = 98 - 80 3d = 18 d = (18)/(3) d = 6 The common difference is 6. Step 3: Substitute the value of d into equation (1) to find the first term a. a + 12(6) = 80 a + 72 = 80 a = 80 - 72 a = 8 The first term is 8. Step 4: Find the last term, which is the 20th term (a_20). Using the formula a_n = a + (n-1)d with n=20, a=8, and d=6: a_20 = 8 + (20-1)6 a_20 = 8 + (19)6 a_20 = 8 + 114 a_20 = 122 The last term is 122.