Question

Find the missing term of this geometric sequence 4,6,_,_,20,25,_

Answer 1

If this is a geometric sequence, then the terms are arⁿ for n≥0.

a=4, ar=6, ar⁴=20, ar⁵=25.

So r=6/4=3/2; r=25/20=5/4. So this is not a geometric sequence because r is not constant.

Let's assume that p(n)=a₀+a₁n+a₂n²+a₆n³. From this we can obtain no more than 4 coefficients because we are given only 4 values. We are assuming that the missing values will fit this function.

p(0)=a₀=4; we can substitute for a₀ for in the following equations:

p(1)=a₀+a₁+a₂+a₃=6; a₁+a₂+a₃=2;

p(4)=a₀+4a₁+16a₂+64a₃=20, 4a₁+16a₂+64a₃=16, a₁+4a₂+16a₃=4;

p(5)=a₀+5a₁+25a₂+125a₃=25; 5a₁+25a₂+125a₃=21, a₁+5a₂+25a₃=4.2.

p(4)-p(1)=3a₂+15a₃=2;

p(5)-p(1)=4a₂+24a₃=2.2.

4(p(4)-p(1))=12a₂+60a₃=8;

3(p(5)-p(1))=12a₂+72a₃=6.6.

Therefore 12a₃=-1.4, a₃=-7/60.

Substituting for a₃, 3a₂-7/4=2, a₂=5/4.

Substituting for a₂ and a₃:

a₁+5/4-7/60=2, a₁=13/15.

p(n)=4+13n/15+5n²/4-7n³/60.

p(2)=9.8, p(3)=14.7, p(6)=29.

One solution is 4,6,9.8,14.7,20,25,29. Although this is a valid solution there are other solutions, but there is insufficient data to find a different solution.