Evaluate the following sums: a) sumk=120 (3k+2) b) sumk=14 (k2+k)

Here are the solutions to the problems. 11. Evaluate the following sums: a) _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 term and apply the sum 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 sum is 670. b) _k=1^4 (k^2+k) Step 1: Use the linearity property of summation to split the sum. _k=1^4 (k^2+k) = _k=1^4 k^2 + _k=1^4 k Step 2: Apply the sum formulas _k=1^n k^2 = (n(n+1)(2n+1))/(6) and _k=1^n k = (n(n+1))/(2). Here n=4. = (4(4+1)(2 × 4+1))/(6) + (4(4+1))/(2) Step 3: Calculate the values. = (4(5)(8+1))/(6) + (4(5))/(2) = (4 × 5 × 9)/(6) + (20)/(2) = (180)/(6) + 10 = 30 + 10 = 40 The sum is 40. 12. The 13^th and 16^th 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: Write down the given information using the formula for the n^th term of an A.P., a_n = a_1 + (n-1)d. For the 13^th term: a_13 = a_1 + (13-1)d a_1 + 12d = 80 (Equation 1) For the 16^th term: a_16 = a_1 + (16-1)d a_1 + 15d = 98 (Equation 2) Step 2: Solve the system of equations to find the common difference d. Subtract Equation 1 from Equation 2: (a_1 + 15d) - (a_1 + 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_1. a_1 + 12(6) = 80 a_1 + 72 = 80 a_1 = 80 - 72 a_1 = 8 The first term is 8. Step 4: Find the last term, which is the 20^th term (a_20), since there are 20 terms. Using the formula a_n = a_1 + (n-1)d with n=20, a_1=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.