In an arithmetic progression (t_3+t_9=70) and (t_5+t_{15}=110). What is the sum of the first (20) terms?

Correct answer: 575

Let the first term be \(a\) and the common difference be \(d\). From \(t_3+t_9=(a+2d)+(a+8d)=70\), we get \(a+5d=35\). Similarly, \(t_5+t_{15}=(a+4d)+(a+14d)=110\) gives \(a+9d=55\). Subtracting gives \(4d=20\), so \(d=5\) and \(a=10\). Therefore, \(S_{20}=\frac{20}{2}[2(10)+19(5)]=575\). The value 1150 can result from incorrectly using \(n\) instead of \(\frac{n}{2}\) in the sum formula. Exam tip: convert the given term relations into equations in \(a\) and \(d\) first.