Question

дуже терміново!
(an) - Aрефметична прогресія, a6=16₁a18=52
треба знайти S30-?

Answer 1

$\displaystyle\bf\\\left \{ {{a_{18} =52} \atop {a_{6} =16}} \right. \\\\\\-\left \{ {{a_{1} +17d=52} \atop {a_{1} +5d=16}} \right. \\---------\\12d=36\\\\d=36:12=3\\\\a_{1}=16-5d=16-5\cdot3=16-15=1\\\\\\S_{30} =\frac{2a_{1} +29d}{2} \cdot 30=(2a_{1} +29d)\cdot 15=\\\\\\=(2\cdot 1+29\cdot 3)\cdot 15=(2+87)\cdot 15=89\cdot 15=1335\\\\\\Otvet \ : \ S_{30}=1335$