A sequence \(T_n\) is given as: \(T_{1}=7;T_{2}=13;T_{3}=19;T_{4}=25\) such that the difference between every two consecutive terms is the same.
Find the value of \(T_{1}+T_{2}+T_{3}+\cdots+T_{50}\).
T50 = T1 + 6(50-1) = 7 + 6*49 = 301
Sum =
(first term + last term) (number of terms)/ 2 =
(7 + 301) (50) / 2 =
(308) * 25 =
7700
