Question

В арифметичнiй прогресiї a4+а11= 27, а10+а6 =15. Обчислiть рiзницю прогресiї.

Answer 1

$\displaystyle\bf\\\left \{ {{a_{4} +a_{11} =27} \atop {a_{10} +a_{6} =15}} \right. \\\\\\\left \{ {{a_{1} +3d+a_{1} +10d=27} \atop {a_{1} +9d+a_{1}+5d =15}} \right. \\\\\\-\left \{ {{2a_{1} +13d=27} \atop {2a_{1} +14d=15}} \right.\\ -----------\\-d=12\\\\\boxed{d=-12}$