Question

In a sequence of four positive numbers, the first three are in geometric progression and the last three are in arithmetic progression. The first number is 12 and the last number is 45/2. The sum of the two middle numbers can be written as a/b where a and b are coprime positive integers. Find a+b

Answer 1

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Answer 2

lets call the common ratio between the 1st three terms 'r', we can then write the sequence of four numbers as:\[12,12r,12r^2,\frac{45}{2}\]you are also told that the last three are in an arithmetic progression, so the difference between the fourth and third numbers must equal the difference between the third and second numbers, i.e.:\[\frac{45}{2}-12r^2=12r^2-12r\]\[\therefore45-24r^2=24r^2-24r\]\[\therefore48r^2-24r-45=0\]divide by 3 to get:\[16r^2-8r-15=0\]factorise to get:\[(4r-5)(4r+8)=0\]\[\therefore r=\frac{5}{4}\qquad\text{or}\qquad r=-2\]the negative value for 'r' does not satisfy the requirement that the last three terms are in an arithmetic progression.\[\therefore r=\frac{5}{4}\]if you substitute this into the sequence of numbers we get:\[12,15,\frac{75}{4},\frac{45}{2}\]you are then told that the sum of the middle two numbers can be expressed as \(\frac{a}{b}\) where 'a' and 'b' are co-prime (i.e. it is a fraction in its lowest terms. therefore:\[\frac{a}{b}=15+\frac{75}{4}=\frac{135}{4}\]that gives you the values for 'a' and 'b'. You should then be able to use this to calculate 'a+b'.