Question

Найти первый член арифм. прогрессии, если:
а1+а5=20
а2+а3=17

Answer 1

$\left \{ {{ a_{1}+ a_{5}=20 } \atop { a_{2}+ a_{3}=17 }} \right.\\\\ \left \{ {{ a_{1}+ a_{1} +4d=20 } \atop { a_{1}+d+ a_{1} +2d=17}} \right. \\\\- \left \{ {{2 a_{1}+4d=20 } \atop {2 a_{1}+3d=17 }} \right.$
_________
d = 3
2a₁ + 4d = 20
a₁ + 2d = 10
a₁ = 10 - 2d = 10 - 2 * 3 = 4
Ответ: a₁ = 4

Answer 2

{а1+а1+4d=20;2a1+4d=20
{a1+d+a1+2d=17;2a1+3d=17
____________________
4d-3d=20-17
d=3
2a1+4d=20
2a1=20-12
a1=4