Question

Помогите пожалуйста! Арифметическая прогрессия.

Answer 1

1)
$a_{n}=a_{1}+(n-1)*d$
$a_{5}=a_{1}+4d \\ a_{9}=a_{1}+8d \\ a_{3}=a_{1}+2d \\ a_{10}=a_{1}+9d \\ \left \{ {a_{1}+4d+a_{1}+8d=42} \atop {(a_{1}+2d)*(a_{1}+9d)=165}} \right.$
$\left \{ {{2a_1+12d=42} \atop {a_{1}^2+11a_{1}d+18d^2=165}} \right. \\ \left \{ {{a_{1}+6d=21} \atop {a_{1}^2+11a_{1}d+18d^2=165}} \right.$
$a_{1}=21-6d \\ (21-6d)^2+11d*(21-6d)+18d^2=165 \\ -12d^2-21d+276=0 \\ 4d^2+7d-92=0 \\ D=7^2-4*4*(-92)=1521$
$d_{1}= \frac{-7+39}{2*4}= \frac{x32}{8}=4 \\ d_{2}= \frac{-7-39}{2*4}=- \frac{46}{8}$ - лишний корень
$a_{1}= 21-6*4=-3$
Ответ: -3; 4
2) 
$S_{n}= \frac{2a_{1}+(n-1)d}{2}*n \\ S_{7}= \frac{2a_{1}+6d}{2}*7 \\ S_{15}= \frac{2a_{1}+14d}{2}*15$
$\left \{ {{ \frac{2a_{1}+6d}{2}*7=94,5 } \atop { \frac{2a_{1}+14d}{2}*15=112,5 }} \right. \\ -\left \{ {{a_1+3d=13,5} \atop {a_{1}+7d=7,5}} \right. \\ -4d=6 \\ d= \frac{6}{-4}=-1,5$$a_{1}=7,5-7*(-1,5)=18$
Ответ: 18; -1,5