В арифметической прогрессии a2+ а6 = 44, а5 а1 = 20. Найдите а100 А) 507;В) 495;C)...

$\left \{ {{a_{1}+d+a_{1}+5d =44} \atop {a_{1}+4d-a_{1} =20}} \right. \\\left \{ {{2a_{1}+6d =44} \atop {4}d =20}} \right. \\$

2a1=44-6d $4(\frac{1}{3}a_{1} -\frac{22}{3})=20\\\frac{4a_{1} }{3}-\frac{88}{3}=20\\ a_{1} =\frac{28}{3}*\frac{3}{4} \\a_{1} =7\\$

-6d=44-2a1

d=1/3a1-22/3

d=7/3-22/3

d=-5

$a_{100}=7-5*99\\a_{100} =502$