If A = 1 + r a + r 2 a + r 3 a + … ∞ and B = 1 + r b + r 2 b + r 3 b + … ∞ , then b a is equal to

A = 1 − r a 1 ⇒ 1 − r a = A 1 ⇒ r a = 1 − A 1 = A A − 1 B = 1 − r b 1 ⇒ 1 − r b = B 1 ⇒ r b = 1 − B 1 = B B − 1 ∴ a lo g r = lo g ( A A − 1 ) and b lo g r = lo g ( B B − 1 ) ∴ b a = l o g ( B B − 1 ) l o g ( A A − 1 ) = lo g B B − 1 ( A A − 1 )

Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesG.P.Easy2 minPYQ_2024

MathematicsEasysingle choice

If $A = 1 + r^{a} + r^{2 a} + r^{3 a} + \dots \infty$ and $B = 1 + r^{b} + r^{2 b} + r^{3 b} + \dots \infty$ , then $\frac{a}{b}$ is equal to

Options:

Answer:

Solution:

$A = \frac{1}{1 - r ^{a}} \Rightarrow 1 - r^{a} = \frac{1}{A} \Rightarrow r^{a} = 1 - \frac{1}{A} = \frac{A - 1}{A}$ $B = \frac{1}{1 - r ^{b}} \Rightarrow 1 - r^{b} = \frac{1}{B} \Rightarrow r^{b} = 1 - \frac{1}{B} = \frac{B - 1}{B}$ $∴ a lo g r = lo g (\frac{A - 1}{A})$ and $b lo g r = lo g (\frac{B - 1}{B})$ $∴ \frac{a}{b} = \frac{l o g ( \frac{A - 1}{A} )}{l o g ( \frac{B - 1}{B} )} = lo g_{\frac{B - 1}{B}} (\frac{A - 1}{A})$

Stream:BITSATSubject:MathematicsTopic:Sequence & SeriesSubtopic:G.P.

⏱ 2mℹ️ Source: PYQ_2024

Doubts & Discussion

Loading discussions...