Let A1, A2, A3,…. be an increasing geometric progression of positive real numbers. If A1A3A5A7 = 1 1 2 9 6 and A2 + A4 = 7 3 6 , then, the value of A6 + A8 + A10 is equal to

Let A 2 A 1 = A 3 A 2 = . . . = r ∴ A 1 A 3 A 5 A 7 = 1 1 2 9 6 ⇒ A 1 r 3 = 1 6 . . . . . . . ( i ) Again, A2 + A4 = 736 A1r=736−16=136........(ii) (i)&(ii)r=6&A1=1366 ∴A6+A8+A10=A1r5(1+r2+r4)=43

Let A₁, A₂, A₃,…. be an increasing geometric progression of positive real numbers. If A₁A₃A₅A₇ = $\frac{1}{1296}$ and A₂ + A₄ = $\frac{7}{36},$ then, the value of A₆ + A₈ + A₁₀ is equal to

Option 1 - 33

Option 2 - 37

Option 3 - 43

Option 4 - 47

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Answered by

Payal Gupta | Contributor-Level 10

7 months ago

Correct Option - 3

Detailed Solution:

Let $\frac{A_{2}}{A_{1}} = \frac{A_{3}}{A_{2}} = . . . = r$

$∴ A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}$

$\Rightarrow A_{1} r^{3} = \frac{1}{6} . . . . . . . (i)$

Again, A₂ + A₄ = $\frac{7}{36}$

$A_{1} r = \frac{7}{36} - \frac{1}{6} = \frac{1}{36} . . . . . . . . (i i)$

$(i) & (i i) r = \sqrt[]{6} & A_{1} = \frac{1}{36 \sqrt[]{6}}$

$∴ A_{6} + A_{8} + A_{10} = A_{1} r^{5} (1 + r^{2} + r^{4}) = 43$

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Kindly consider the following figure

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