What is the 55th term of a sequence if the 6th term is 4 and the 100th term is 98?

What is the 55th term of a sequence if the 6th term is 4 and the 100th term is 98 ?

$$\small{
\begin{array}{rcl}
a_n &=& a_1\cdot r^{n-1} \\
a_6 &=& a_1\cdot r^5 = 4 \\
a_{100} &=& a_1\cdot r^{99} = 98\\
\\
\hline
\\
\dfrac{a_{100}}{a_6} &=&
\dfrac{a_1\cdot r^{99}}{a_1\cdot r^{5}} = \dfrac{98}{4} \\\\
\dfrac{a_1\cdot r^{99}}{a_1\cdot r^{5}} &=& \dfrac{98}{4} \\\\
\dfrac{r^{99}}{r^{5}} &=& 24.5 \\\\
r^{99-5} &=& 24.5 \\\\
r^{94} &=& 24.5 \\\\
r &=& \sqrt[94]{24.5} \\\\
r &=& 1.03461402806 \\
\\
\hline
\\
a_6 &=& a_1\cdot r^5 = 4 \\\\
a_1 \cdot r^{5} &=& 4 \\\\
a_1 &=& \dfrac{4} { r^5 } \\\\
a_1 &=& \dfrac{4} { 1.03461402806^5 } \\\\
a_1 &=& 3.37417946747\\
\\
\hline
\\
a_{55} &=& a_1\cdot r^{54} \\\\
a_{55} &=& 3.37417946747 \cdot 1.03461402806^{54}\\\\
a_{55} &=& 21.1933572378
\end{array}
}$$

If this is an arithmetic sequence then the n'th term is a + (n-1)d where a is the first and d is the difference.

a + 5d = 4

a + 99d = 98

so a = -1 and d = 1

the 55th term is -1 + 54*1 = 53

.

If this is a geometric sequence then the n'th term is a*r^n-1 where r is the multiplier.

a*r⁵ = 4

a*r⁹⁹ = 98

r=(98/4)^1/94 ≈ 1.0346

a = 4/1.0346⁵ ≈ 3.3742

The 55th term is 3.3742*1.0346⁵⁴ ≈ 29.193

.