∵等差数列{an}的前n项和为Sn，且S4≥10，S5≤15∴S4＝4a1+4×32d≥10S5＝5a1+5×42d≤15即2a1+3d≥5a1+2d≤3∴a4＝a1+3d≥5−3d2+3d＝5+3d2a4＝a1+3d＝(a1+2d)+d≤3+d∴5+3d2≤a4≤3+d，5+3d≤6+2d，d≤1∴a4≤3+d...