If a1,a2,a3… and b1,b2,b3… are geometric sequences, show that a1b1,a2b2,a3b3,… is also geometric sequence.

Use the geometric sequence property If a,,b and c are geometric sequence then b2=a×c Now a1,a2,a3… are in G.P. a22=a1×a3 Also b1,b2,b3… are in G.P. ⇒b22=b1×b3 Now the sequence a1b1,a2b2,a3b3,… would be in G.P If (a2b2)2=a1b1×a3b3 Consider the left-hand side (a2b2)2 Simplifying, we get (a2b2)2=a22×b22 Substitute a22=a1×a3 and ⇒b22=b1×b3 in above equation ⇒(a2b2)2=a1×a3×b1×b3 Rewrite the above equation (a2b2)2=(a)1b1a3b3 Which implies that a1b1,a2b2,a3b3,… are in a geometric sequence

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Answered question

2021-08-21

If $a_{1}, a_{2}, a_{3} \dots and b_{1}, b_{2}, b_{3} \dots$ are geometric sequences, show that $a_{1} b_{1}, a_{2} b_{2}, a_{3} b_{3}, \dots$ is also geometric sequence.

Answer & Explanation

BleabyinfibiaG

Skilled2021-08-22Added 118 answers

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