Question

For the G.P.s
i) If a = 7, r = 1/3, find $t_{6}$.
ii) If a = 5, r = -2, find $t_{5}$.
iii) If a = 2/3, $t_{6}$ = 162, find r.
iv) If t$t_{8}$ = 640, r = 2, find a.

Answer 1

Answer:

i) 7/243

ii)80

iii)3

iv)20

Step-by-step explanation:

Hi,

We know that in a geometric progression Tn is

given by a*rⁿ⁻¹, where a is first term and r common

ratio.

i) Given a = 7 and r = 1/3

T₆ = a*r⁵

= 7*(1/3)⁵

= 7/3⁵

= 7/243

ii) i) Given a = 5 and r = -2

T₅ = a*r⁴

= 5*(-2)⁴

= 5*16

= 80

iii) Given a = 2/3 and T₆ = 162

We know T₆ = a*r⁵

= 2/3*(r)⁵ = 162

(r)⁵ = 243

r = 3

iv) Given r = 2 and T₈ = 640

We know T₈ = a*r⁷

= a*(2)⁵ = 640

32a = 640

a = 20

Hope, it helps !

Answer 2

$\textsf \green { Solution }$

$\textsf \blue {We know that , nth term in G.P}$

$\red t_{n} = a \cdot r^{n-1}$

i ) Given a = 7 , r = 1/3 ,

$t_{6} = ?$

$t_{6} = 7 \cdot \left ( \frac {1}{3} \right )^{5}$

= $ 7 × 1/3^{5}$

= $7/243$

ii ) If a = 5 , r = -2 ,

find $t_{5}$

$t_{5} = 5 \cdot \left ( - 2 \right )^{4}$

= $ 5 × 16 $

= $80$

iii )) If a = 2/3, $t_{6}$ = 162, find r.

$\frac {2}{3} \cdot r^{n-1} = 162$

$\implies r^{6-1} = 162 \cdot \frac{3}{2}$

$\implies r^{5} = 3^{5}$

$r = 3$

iv) If t$t_{8}$ = 640, r = 2, find a.

$a \cdot 2^{7} = 640$

$\implies a = \frac {640}{128}$

$\implies a = 5$

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