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

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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

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