Question

Сумма 3 и 9 арифметической прогрессии равна 6,а их произведения равны 135/16. Найти первые 15 члены прогрессии.

Answer 1

$left { {{ a_{3}+ a_{9} =6 } atop { a_{3} * a_{9} = frac{135}{16} }} right. \ \ left { {{ a_{1} +2d+a_{1} +8d=6} atop {(a_{1} +2d)(a_{1} +8d)= frac{135}{16} }} right. \ \ left { {{ a_{1}=3-5d} atop {(a_{1} +2d)(a_{1} +8d)= frac{135}{16} }} right. \ \ left { {{ a_{1}=3-5d} atop {(3-5d+2d)(3-5d+8d)= frac{135}{16} }} right. \ \ left { {{ a_{1}=3-5d} atop {(3-3d)(3+3d)= frac{135}{16} }} right. \ \ left { {{ a_{1}=3-5d} atop {9-9d^2= frac{135}{16} }}$
$left { {{ a_{1}=3-5d } atop {1-d^2= frac{15}{16} }} right. \ \ left { {{ a_{1}=3-5d } atop {16d^2=1 }} right. \ \ d=pm frac{1}{4} \ \ 1)d= frac{1}{4}=0.25 ; \ a_{1} =3- frac{5}{4} = frac{7}{4} =1.75; \ a_{15} =1.75+0.25*14=5.25; \ S_{15} = frac{1.75+5.25}{2} *15=3.5*15=52.5 \ \ 2)d=- frac{1}{4}=-0.25 ; \ a_{1} =3+ frac{5}{4} = frac{17}{4} =4.25; \ a_{15} =4.25-0.25*14=0.75; \ S_{15} = frac{4.25+0.75}{2} *15=2.5*15=37.5$