Question

дано: с4=40, с15=12, найти д?​

Answer 1

$\left \{ {{c_{15}=12 } \atop {c_{4}=40 }} \right. \\\\-\left \{ {{c_{1}+14d=12 } \atop {c_{1}+3d=40}} \right.\\-------\\ 11d=-28\\\\d=-2\frac{6}{11}$