Question

the 5th, 8th and 11th terms of a GP are a, b, c respectively. Show that b^2 = ac.

guys please help​

Answer 1

So we have,

$AR^4=a\\\\AR^7=b\\\\AR^{10}=c$

where $A\ \&\ R$ are first term and common ratio of the GP respectively.

We have to prove that $b^2=ac.$

Taking LHS,

$b^2=(AR^7)^2\\\\b^2=A^2R^{14}\\\\b^2=A^{1+1}R^{4+10}\\\\b^2=AR^4\cdot AR^{10}\\\\b^2=ac$

Thus we've arrived at the RHS.

Since $b^2=ac,$ we can say that $a,\ b,\ c$ form a GP whose common ratio is $R^3.$

Well, the following also proves that $b^2=ac,$ is true.

$\dfrac {b}{a}=\dfrac {c}{b}\\\\\\\dfrac {AR^7}{AR^4}=\dfrac {AR^{10}}{AR^7}=R^3$

QED

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Well it is true that, at least three terms in a non - constant GP, among which the consecutive terms are equidistant, also form a GP. Let's prove this statement in the case of three terms.

Suppose there exists three terms T_(n - d + 1), T_(n + 1) and T_(n + d + 1) in a GP. Let,

$T_{n-d+1}=a=AR^{n-d}\\\\T_{n+1}=b=AR^n\\\\T_{n+d+1}=c=AR^{n+d}$

Then we see that,

$\dfrac {AR^n}{AR^{n-d}}=\dfrac {AR^{n+d}}{AR^n}=R^d\\\\\\\therefore\ b^2=ac\quad\iff\quad\dfrac {b}{a}=\dfrac {c}{b}$

Thus they also form a GP.

#answerwithquality

#BAL