# SOLUTION: Tuition: A student paid \$6500 tuition for her first year attending a university. Tuition at the university is projected to increase \$750 a year for the next four years. (a) H

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 Click here to see ALL problems on decimal-numbers Question 85945: Tuition: A student paid \$6500 tuition for her first year attending a university. Tuition at the university is projected to increase \$750 a year for the next four years. (a) How much tuition should the student expect to pay for her fourth year at the university? (b) How much should the student expect to pay in tuition for all four years? TAnswer by bucky(2189)   (Show Source): You can put this solution on YOUR website!First year tuition = \$6500 Second year tuition = \$6500 + \$750 = \$7250 Third year tuition = \$7250 + \$750 = \$8000 Fourth year tuition = \$8000 + \$750 = \$8750 Graduation (a) She should expect to pay \$8750 for her fourth year tuition. (b) The total tuition she will pay is \$6500 + \$7250 + \$8000 + 8750 = \$30,500 . You might notice that this is an arithmetic progression. The first term (call it "a") is \$6500. The common difference between terms (call it "d") is \$750. The number of terms in the progression (call it "n") is 4, and the last term (call it "L") is 4. . The last term in an arithmetic progression is given by the equation: . L = a + (n-1)*d . Substitute the known values for a, n, and d to get: . L = \$6500 + (4-1)*\$750 = \$6500 + 3*\$750 = \$6500 + \$2250 = \$8750 . Notice that this is the answer to question (a) above. She should expect to pay \$8750 in tuition for her last year. . The sum (call it "S")of a finite number of terms in an arithmetic progression is given by the equation: . . Substitute the values of a, d, and n for this problem and the sum becomes: . . So using the equation for the sum you can find the total paid for the four years of tuition is \$30,500 ... the same as we calculated previously. . Hope this helps you to understand arithmetic progressions and the equations for finding the last term and the sum of the terms.