В арифметической прогрессии

S2-S4+a2=14;

S3+a3=17

Нужно найти a1, d

A1+a1+d - [ (a1+a1+3d) * 4/2]+a1+d =

= 3a1+2d-4a1-6d=-a1-4d=14 a1=-4d-14

(a1+a1+2d) * 3/2+a1+2d = 3a1+3d+a1+2d=

4a1+5d=17 4 (-4d-14) + 5d=-16d-56+5d=17

-11d=73 d=-73/11 = - 6 7/11

a1=4*73/11-14 = 292/11-14 = 26 6/11-14=12 6/11