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Algebra of Limits

Let a1, a2,...

Question

Let $a_{1}, a_{2}, . . . . . .$ be a G.P. such that $\frac{a_{4}}{a_{6}} = \frac{1}{4}$ and $a_{2} + a_{5} = 216$ , then $a_{1}$ equals

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Solution

in GP $a_{n} = a_{1} \times r^{(n - 1)}$
$\frac{a_{4}}{a_{6}} = \frac{1}{r^{2}} = \frac{1}{4}$
$⟹ r = 2$
$a_{2} + a_{5} = a_{1} (r + r^{4}) = a_{1} (18) = 216$
$⟹ a_{1} = 12$

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Similar questions

a_{n}

be a G.P such that

\frac{a_{4}}{a_{6}} = \frac{1}{4}

a_{2} + a_{5} = 216

a_{1} =

a_{1}, a_{2}, . . . . . . . . . . . . ., a_{50}

G . P .

\frac{a_{1} - a_{3} + a_{5} - . . . . . + a_{49}}{a_{2} - a_{4} + a_{6} - . . . . . + a_{50}}

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} a n d a_{6}

are consecutive odd natural numbers such that

a_{6} > a_{1}, a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} = p^{3} a n d a_{5} = 39

, then the value of

(a_{3} + a_{4} - 12 p)

a_{1}, a_{2}, a_{3} a_{4}, a_{5}

G . P

of positive real numbers such that the

A . M

a_{2}

a_{4}

117

G . M

a_{2}

a_{4}

108

A . M

a_{1}

a_{5}

a_{1}, a_{2}, a_{3}, a_{4}, a_{5}

be a G.P. of positive real numbers such that the A.M. of

a_{2}

a_{4}

117

and the G.M. of

a_{2}

a_{4}

108

. Then the A.M. of

a_{1}

a_{5}

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