Question

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

Answer 1

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