Question

The second term is 8 and the fifth term is -27. Find a1 in this geometric sequence.

Answer 1

`a_(1) = (-16)/(3)`

Now the Geometric sequence

`(-16)/(3) , 8 ,-12 ,18 ,-27`

Explanation:

Step(i):-

Given second term is '8'

`t_(n)= a r^(n-1)`

`t_(2)= a r^(2-1)`

`a r^(2-1) = 8`

a r = 8 ...(i)

And also given fifth term is -27

`t_(5)= a r^(5-1)`

`a r^(4) = -27` ...(ii)

Step(ii):-

Now simplify

`(a r^(4) )/(ar) = (-27)/(8)`

`r^(3) = (-27)/(8)`

`r^3 =((-1)^(3) (3)^(3) )/((2)^(3) )`

`(r^3)^{(1)/(3) } =(((-1)^(3) (3)^(3) )/((2)^(3) ))^{(1)/(3) }`

On simplification, we get

`r = (-3)/(2)`

now from(i)

a r = 8

`a((-3)/(2) ) = 8`

`a = (-16)/(3)`

`a_(1) = (-16)/(3)`

Now the Geometric sequence

`(-16)/(3) , 8 ,-12 ,18 ,-27`