Chapter 11 - Further Topics in Algebra - Chapter 11 Test Prep - Review Exercises - Page 1081: 7

$$4,\,\,\,4.5,\,\,\,5,\,\,\,5.5,\,\,\,6$$

$$\eqalign{ & {\text{Five terms, then }}n = 5 \cr & {\text{First term }}{a_1} = 4 \cr & {\text{The sum of the five terms equal to 25,}}\,\,\,{\text{then }}{S_5} = 25 \cr & \cr & {\text{The Sum of the First }}n{\text{ Terms is }} \cr & {S_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right] \cr & n = 5 \cr & {S_5} = \frac{5}{2}\left[ {2{a_1} + \left( {5 - 1} \right)d} \right] \cr & 25 = \frac{5}{2}\left[ {2\left( 4 \right) + 4d} \right] \cr & 10 = 2\left( 4 \right) + 4d \cr & 2 = 4d \cr & d = \frac{1}{2} \cr & \cr & {\text{The }}nth{\text{ term is}} \cr & {a_n} = {a_1} + \left( {n - 1} \right)d \cr & {a_n} = 4 + \left( {n - 1} \right)\left( {\frac{1}{2}} \right) \cr & {a_n} = 4 + \frac{1}{2}n - \frac{1}{2} \cr & {a_n} = \frac{1}{2}n + \frac{7}{2} \cr & \cr & {a_2} = \frac{1}{2}\left( 2 \right) + \frac{7}{2} = 4.5 \cr & {a_3} = \frac{1}{2}\left( 2 \right) + \frac{7}{2} = 5 \cr & {a_4} = \frac{1}{2}\left( 2 \right) + \frac{7}{2} = 5.5 \cr & {a_5} = \frac{1}{2}\left( 2 \right) + \frac{7}{2} = 6 \cr & \cr & 4,\,\,\,4.5,\,\,\,5,\,\,\,5.5,\,\,\,6 \cr} $$