Question

Question #cc266

Answer 1

`108, 72, 48`

Explanation:

To write the first three terms, we need to find `a` and `r`

We have `T_3 = 48 and T_6 = 14 2/9`

The terms can be written as:

`T_3 = ar^2 and T_6 = ar^5`

Divide the two terms:

`T_6/T_3 = (ar^5)/(ar^2) = (14 2/9)/48`

Simplifying leads to:

`(cancelar^5)/(cancelar^2) = (128/9)/(48/1)`

`r^3 = 128/(48xx9)`

`r^3 = 8/27`

`r = root3(8/27) = 2/3`

You could substitute to find `a`, or realise that that in a G.P. any next term is found by multiplying by the common ratio, `(r)`

`T_1 xx r " "rarr T_2 xx r " "rarr T_3`

As we have the third term `(14 2/9),` we can find the preceding terms by dividing by `r`

`T_3 = 48`

`T_2 =48 div 2/3 = 48 xx3/2 =72`

`T_1 = a = 72 div 2/3 = 72 xx 3/2 = 108`

Check: `T_6 = 108 xx (2/3)^5 = 14 2/9`

Answer 2

`108,72,48,32`

Explanation:

`"the "color(blue)"nth term of a geometric sequence"` is.

`•color(white)(x)a_n=ar^(n-1)`

`"where a is the first term and r the common ratio"`

`"to obtain the first 4 terms we require to find a and r"`

`a_3=ar^2=48to(1)`

`a_6=ar^5=128/9to(2)`

`"divide equation "(2)" by equation "(1)`

`rArr(cancel(a)r^5)/(cancel(a)r^2)=128/9xx1/48=8/27`

`rArrr^3=8/27rArrr=root(3)(8/27)=2/3`

`"substitute "r=2/3" in equation "(1)`

`rArra=48xx9/4=108`

`"to obtain each term in the sequence multiply the"`
`"previous term by r"`

`rArra_1=108`

`rArra_2=108xx2/3=72`

`rArra_3=72xx2/3=48`

`rArra_4=48xx2/3=32`