Question

[Hanuman]

giúp tớ vsssssss

Tìm số hạng đầu và công sai

Answer 1

[Nguyễn Lê Phước Thịnh]

h: Đặt u1=a; q=b

Theo đề, ta có hệ:

\(\left\{{}\begin{matrix}a+a+b+a+2b=9\\a\left(a+b\right)\left(a+2b\right)=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=9\\a\left(a+b\right)\left(a+2b\right)=15\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}b=3-a\\a\left(a+3-a\right)\left(a+6-2a\right)=15\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}b=3-a\\3a\left(-a+6\right)=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=3-a\\-a^2+6a-5=0\end{matrix}\right.\Leftrightarrow\left(a,b\right)\in\left\{\left(1;2\right);\left(5;-2\right)\right\}\)

i: Đặt u1=a; q=b

Theo đề, ta có hệ:

\(\left\{{}\begin{matrix}a+a+b+a+2b+a+3b=20\\a\left(a+b\right)\left(a+2b\right)\left(a+3b\right)=384\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4a+6b=20\\a\left(a+b\right)\left(a+2b\right)\left(a+3b\right)=384\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2a+3b=10\\a\left(a+b\right)\left(a+2b\right)\left(a+3b\right)=384\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{10-3b}{2}\\\dfrac{10-3b}{2}\cdot\dfrac{10-3b+2b}{2}\cdot\dfrac{10-3b+4b}{2}\cdot\dfrac{10-3b+6b}{2}=384\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=5-1.5b\\\left(10-3b\right)\left(10-b\right)\left(10+b\right)\left(10+3b\right)=6144\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=5-1.5b\\\left(100-9b^2\right)\left(100-b^2\right)=6144\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=5-1.5b\\10000-1000b^2+9b^4-6144=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b^2\in\left\{\dfrac{964}{9};4\right\}\\a=5-1.5b\end{matrix}\right.\)

TH1: b^2=964/9

=>\(b\in\left\{\dfrac{2\sqrt{241}}{3};\dfrac{-2\sqrt{241}}{3}\right\}\)

=>\(\left[{}\begin{matrix}a=5-\sqrt{241}\\a=5+\sqrt{241}\end{matrix}\right.\)

TH2: b^2=4

=>b=2 hoặc b=-2

=>a=5-1,5b=5-1,5*2=5-3=2 hoặc a=5-1,5b=5+3=8