Let us assume the sequence to be a1,a2,a3,...a100 According to the question: a1+a2+...a6=a2+a3+...a7 ⇒ a1=a7. Similarly we can prove that a2=a8, a3=a9 and so on. Combining these equations we get ak=ak+6=ak+12= .....for 3,1 natural numbers k. Therefore a19=a25=....=a55=55 Also a33=a39=....=a99=33 So the required sum is 55 + 33 = 88.