Question

Знайдіть суму шести перших членів арифметичної прогресії (bn), якщo a1+a3=12 i a2+a4=16

Answer 1

$\displaystyle\bf\\\left \{ {{a_{1} + a_{3}=12 } \atop {a_{2} + a_{4}=16 }} \right. \\\\\\\left \{ {{a_{1} + a_{1}+2d=12 } \atop {a_{1} +d+ a_{1}+3d=16 }} \right. \\\\\\\left \{ {{2a_{1} + 2d=12 } \atop {2a_{1} + 4d=16 }} \right. \\\\\\-\left \{ {{a_{1} + 2d=8 } \atop {a_{1} + d=6 }} \right. \\--------\\d=2\\\\a_{1} =6-d=6-2=4\\\\\\S_{6} =\frac{2a_{1} +5d}{2} \cdot 6=(2a_{1}+5d)\cdot 3=(2\cdot 4+5\cdot 2)\cdot 3=\\\\\\=(8+10)\cdot 3=18\cdot 3=54$