[tex]\displaystyle a).10+20+30+...+1230=10(1+2+3+...+123)= \\ =10\cdot \frac{123(123+1)}{2} =10 \cdot \frac{123 \cdot 124}{2} =10 \cdot \frac{15253}{2} =10 \cdor 7626=76260 \\ \\ b).7+14+21+...+315=7(1+2+3+...+45)= \\ \\ =7 \cdot \frac{45(45+1)}{2} =7 \cdot \frac{45 \cdot 46}{2} =7 \cdot \frac{2070}{2} =7 \cdot 1035=7245[/tex]
