\(\displaystyle Before \ you \ panic, \ take \ a \ deep \ breath \ and \ relax.\)
\(\displaystyle Now, \ assume \ you \ know \ nothing \ about \ progressions.\)
\(\displaystyle From \ the \ problem: \ Let \ a_1 \ = \ 100, \ then \ a_2 \ = \ 125, \ a_3 \ =150, \ etc.\)
\(\displaystyle or \ a_1 \ = \ 100+(0)(25)\)
\(\displaystyle \ \ a_2 \ = \ 100+(1)(25)\)
\(\displaystyle \ \ a_3 \ = \ 100+(2)(25)\)
\(\displaystyle \ \ a_n \ = \ 100+(n-1)(25)\)
\(\displaystyle a_n \ = \ a_1+(n-1)(25), \ a_1 \ = \ 100, \ n \ = \ number \ of \ terms.\)
\(\displaystyle Now, let \ 25 \ = \ d, \ d \ being \ the \ difference \ of \ two \ consecutive \ terms, \ d \ is \ positive\)
\(\displaystyle if \ the \ terms \ are \ increasing, \ negative \ if \ the \ terms \ are \ decreasing.\)
\(\displaystyle Hence, \ a_n \ = \ a_1+(n-1)d \ and \ we \ call \ this \ equation \ an \ arithmetic \ progression.\)
\(\displaystyle Now, \ we \ want \ to \ know \ how \ much \ it \ would \ cost \ to \ build \ a \ 90 \ ft. \ (n=9) \ tower.\)
\(\displaystyle Sum \ = \ a+(a+d)+(a+2d)+...+(l-2d)+(l-d)+l, \ a \ = \ first \ term, \ l \ = \ last \ term.\)
\(\displaystyle Sum \ = \ l+(l-d)+(l-2d)+...+(a+2d)+(a+d)+a, \ (sum \ is \ commutative).\)
\(\displaystyle Hence, adding, \ we \ get \ 2S \ = \ (a+l)+(a+l)+(a+l)+...+(a+l)+(a+l)+(a+l) \ = \ n(a+l)\)
\(\displaystyle Ergo, \ S \ = \ \frac{n}{2}(a+l) \ and \ last \ term \ (l) \ = \ a_1+(n-1)d, \ (n=9), \ (a=a_1).\)
\(\displaystyle Therefore, \ l \ = \ 100+(8)(25) \ = \ 300, \ \implies \ S \ = \ \frac{9}{2}(100+300) \ = \ \$1800.00\)
\(\displaystyle Note: \ We \ could \ have \ just \ added \ up \ the \ nine \ terms \ for \ the \ answer, \ but \ what \ if\)
\(\displaystyle we \ had \ 900 \ or \ 9000 \ terms?\)
