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Geometric Progression

In a G.P. if ...

Question

In a G.P. if $t_{3} = 2$ and $t_{6} = - \frac{1}{4},$ then $t_{10} =$

A

- \frac{1}{128}

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B

- \frac{1}{64}

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C

\frac{1}{128}

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D

\frac{1}{64}

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Solution

The correct option is B $- \frac{1}{64}$
$t_{3} = 2, t_{6} = - \frac{1}{4}$
$a r^{2} = 2, a r^{5} = - \frac{1}{4}$
$\Rightarrow \frac{1}{r^{3}} = - 8$
$\Rightarrow r = - \frac{1}{2}$
$\Rightarrow a = 8$
$t_{10} = a r^{9} = 8 {(\frac{- 1}{2})}^{9} = \frac{- 1}{64}$

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Similar questions

Q. In a G.P. if

t_{3} = 2

t_{6} = - \frac{1}{4},

t_{10} =

t_{n}

n^{t h}

term of a G.P. If

t_{3} = 2

t_{6} = - \frac{1}{4},

then the value of

t_{10}

t_{n}

n^{t h}

term of a G.P. If

t_{3} = 2

t_{6} = - \frac{1}{4},

then the value of

t_{10}

Find the G.P. if

(a) T₃ : T₆ = 1 : 8 and T₅ = 64

(b) T₄ : T8 = 1 : 81 and T₆ = 486

(c) S₁₀ : S₅ = 33 : 1 and T₆= 32

(d) S₈ : S₄ = 97 : 81 and T₅ = 16

T_{2} + T_{5} = 216

T_{4} : T_{6} = 1 : 4

and all terms are integers, then its first term is:

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Geometric Progression

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