Given an arithmetic progression with Asub50=10 and Asub100 = -10. Find d and Ssub50

Question

Progression, Arithmetic progression

Answer 1

Between the two terms the total difference, where d=difference between consecutive terms, is (100-50)d=-10-10=-20. So d=-20/50=-2/5=-0.4.

The difference between Asub50 and Asub0 is the same as between Asub100 and Asub50, that is, -20, so Asub50-Asub0=-20; 10-Asub0=-20, and Asub0=30, the first term. For example, Asub1=30-2/5=29.6, Asub2=29.2.

The sum Ssub50=

30+29.6+29.2+...=30+30-0.4+30-0.8+30-1.2+...=51*30-0.4(1+2+3+...+50)=

1530-0.4(50*51/2)=1020.

[SsubN=30(N+1)-0.4(N(N+1)/2) where N starts at 0]