A quicker route to the same seven numbers is to first solve the puzzle in "ratio units," ignoring the actual size of D, and only bring in E = 960 at the very end.
Assume for a moment that D = 1 (any positive starting value works, since the whole picture just scales up or down together). The seven numbers sit in the order D, C, B, A, G, F, E, with D and E at the two ends, and A/D = G/C = F/A all equal to one number k, where A is also the fourth term of the A.P. and the first term of the G.P.
So in ratio units: D = 1, C = 2, B = 3, A = 4, G = 8, F = 16, E = 32.
Before scaling up, it is worth checking that these seven ratio-unit numbers already satisfy every condition in the directions, since scaling by a positive constant never breaks an order relation, a common difference/ratio relation, or a ratio like A/D:
Every condition holds in ratio units, so scaling the whole picture by a single positive number keeps every condition true. Now scale everything so E matches the real value 960. Since E = 32 in ratio units and E = 960 in real units, the scale factor is 960/32 = 30. Multiplying every ratio-unit value by 30 gives:
\[ D=30,\ C=60,\ B=90,\ A=120,\ G=240,\ F=480,\ E=960 \]These match the direct method exactly, and since every condition already held true in ratio units, it automatically still holds true after scaling by 30.
D, in real units, is 30.
Let's summarize:
\[ 5m \sum_{r=m}^{2m} T_r \text{ is equal to:} \]