Wierzycaz

2021-08-21

If are geometric sequences, show that ${a}_{1}{b}_{1},{a}_{2}{b}_{2},{a}_{3}{b}_{3},\dots$ is also geometric sequence.

BleabyinfibiaG

Use the geometric sequence property
If a,,b and c are geometric sequence then ${b}^{2}=a×c$
Now ${a}_{1},{a}_{2},{a}_{3}\dots$ are in G.P.
${a}_{2}^{2}={a}_{1}×{a}_{3}$
Also ${b}_{1},{b}_{2},{b}_{3}\dots$ are in G.P.
$⇒{b}_{2}^{2}={b}_{1}×{b}_{3}$
Now the sequence ${a}_{1}{b}_{1},{a}_{2}{b}_{2},{a}_{3}{b}_{3},\dots$ would be in G.P
If ${\left({a}_{2}{b}_{2}\right)}^{2}={a}_{1}{b}_{1}×{a}_{3}{b}_{3}$
Consider the left-hand side ${\left({a}_{2}{b}_{2}\right)}^{2}$
Simplifying, we get
${\left({a}_{2}{b}_{2}\right)}^{2}={a}_{2}^{2}×{b}_{2}^{2}$
Substitute ${a}_{2}^{2}={a}_{1}×{a}_{3}$ and $⇒{b}_{2}^{2}={b}_{1}×{b}_{3}$ in above equation
$⇒{\left({a}_{2}{b}_{2}\right)}^{2}={a}_{1}×{a}_{3}×{b}_{1}×{b}_{3}$
Rewrite the above equation
$\left({a}_{2}{b}_{2}{\right)}^{2}=\left(a{\right)}_{1}{b}_{1}{a}_{3}{b}_{3}$
Which implies that ${a}_{1}{b}_{1},{a}_{2}{b}_{2},{a}_{3}{b}_{3},\dots$ are in a geometric sequence

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