Question

In a G.P if t6=-(1/32), t9=1/256 then find the11 so I will mark you as brainliest

Answer 1

Let a be the first term and r be the the common ratio
then the geometric series is a, ar, ar².....

tn = a. r ^(n-1)

t6 = a. r^5 = -1/32 »»»»»●

t9 = a. r^8= 1/256

= a. r^(5+5) = a. r^5. r^3= 1/256

from ●
we get

-1/32. r^3 = 1/256

r^3 = -32/256 = -1/8

r = -1/2

by substituting in ●

we get, a = -2^5/(-32) = 1

so, t11 = a.r^10 = 1. (-1/2)^10 = 1/1024

is your answer

Hope you understand

any doubts, comment ;)(

Answer 2

Answer:

t11=1/1024

Step-by-step explanation:

Sixth term = T6 = -1/32

ninth term = T9 = 1/256

then , common ratio = R = ?

11th term = T11 = ?

_______________

R^p-q = Tp/Tq

R^9-6 = (1/256)/-1/32

R^3 = 1/256 × -32

R^3 = 1/-8

R^3 = -1/8

R^3 = (-1/2)^3

R = -1/2 .

__________

T6 = a(R)^n-1

-1/32 = aR^6-1

-1/32 = aR^5

-1/32 = a × (-1/2)^5

-1/32 = a × -1/32

(-1/32)/(-1/32) = a

-1/32 × -32 = a

-1×-1 = a

1 = a

a = 1

________________

T11 = aR^n-1

T11 = (1)(-1/2)^11-1

T11 = (-1/2)^10

T11=1/1024

_________________