If you could write out the whole problem exactly how it is, I can solve it for you.
their is no answer
to that problem
50 Is the answer.........
Answer:
We are effectively looking for a and b such that 5, a, b, 135 is a geometric sequence.
This sequence has common ratio <span><span>3<span>√<span>1355</span></span></span>=3</span>, hence <span>a=15</span> and <span>b=45</span>
Explanation:
In a geometric sequence, each intermediate term is the geometric mean of the term before it and the term after it.
So we want to find a and b such that 5, a, b, 135 is a geometric sequence.
If the common ratio is r then:
<span><span>a=5r</span><span>b=ar=5<span>r2</span></span><span>135=br=5<span>r3</span></span></span>
Hence <span><span>r3</span>=<span>1355</span>=27</span>, so <span>r=<span>3<span>√27</span></span>=3</span>
Then <span>a=5r=15</span> and <span>b=ar=15⋅3=45</span>
Answer:
Ok so I’m stuck with this question too But I got the top part.
p=1/6
q=2/6
r=3/6
s=3/6
