In an increasing geometric progression of positive terms, the sum of the second and sixth terms is and the product of the third and fifth terms is 49. Then the sum of the 4th ,6th and 8th terms is equal to:

Engineering

Mathematics

Introduction to Determinants

Question

In an increasing geometric progression of positive terms, the sum of the second and sixth terms is $\frac{70}{3}$ and the product of the third and fifth terms is 49. Then the sum of the 4^th ,6^th and 8^thterms is equal to:

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T₂ + T₆= 70/3
ar + ar⁵ = 70/3 ⇒ ar (1 + r⁴) = 70/3 __________(1)
T₃ × T₅ = 49
ar². ar⁴ = 49 ⇒ (ar)² r⁴ = 49 __________(2)

from (1) and (2) $\frac{{(1 + r^{4})}^{2}}{r^{4}} = \frac{70 \times 70}{3 \times 3 \times 49}$

$(r^{2} + \frac{1}{r^{2}}) = \frac{10}{3} = 3 + \frac{1}{3} r^{r} = 3, r = + \sqrt{3}, a = \frac{7}{3 \sqrt{3}}$

T₄ + T₆ + T₈ = a(r³ + r⁵ + r⁷)

$= \frac{7}{3 \sqrt{3}} (3 \sqrt{3} + 9 \sqrt{3} + 27 \sqrt{3}) = 91$

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