t3l+1=1197. The terms are t1,t4,t7,...,t55:19 terms, A.P. with first term t1=a, common difference 3d. ∑=‌
19
2
[2a+(19−1)⋅3d]=1197 ‌
19
2
(2a+54d)=1197⇒2a+54d=126⇒a+27d=63 2. Use t7+3t22=174. t7=a+6d,‌‌t22=a+21d (a+6d)+3(a+21d)=174 a+6d+3a+63d=174⇒4a+69d=174 From (1): a=63−27d. Substitute in (2): 4(63−27d)+69d=174 252−108d+69d=174⇒252−39d=174⇒d=2 a=63−27⋅2=9 So tn=9+(n−1)⋅2 First 9 terms: 9,11,13,15,17,19,21,23,25. Sum of squares: