Find the 2015th term of the geometric sequence 1,−1/3,-1/6,−1/9 ,.
+26375 Find the 2015th term of the geometric sequence 1,−1/3,-1/6,−1/9 ,.
In a geometric sequence, the common ratio, r between any two consecutive terms is always the same.
\(\begin{array}{rcl} \dfrac{ \text{Any term} } { \text{Previous term}} &=& \dfrac{u_n}{ u_n-1} = \text{ constant } = r \\ \end{array}\)
\(\begin{array}{rcr} 1,-\frac13,-\frac16,-\frac19,-\cdots \\ u_1 &=& 1\\ u_2 &=& -\frac13\\ u_3 &=& -\frac16\\ u_4 &=& -\frac19\\ \cdots \\ \end{array} \)
If three terms, \(u_n,\ u_{n+1},\ u_{n+2}\) are in geometric sequence, then:
\(\begin{array}{rcl} \dfrac{ u_{n+2} } { u_{n+1} } &=& \dfrac{ u_{n+1} } { u_{n} }\\ \end{array}\)
We have
\(\begin{array}{rcr} \dfrac{ u_2 } { u_1 } = \dfrac{ -\frac13 } { 1 } = -\frac13 &\ne& \dfrac{ u_3 } { u_2 } = \dfrac{ -\frac16 } { -\frac13 } = \frac12 \\ &\ne& \dfrac{ u_4 } { u_3 } = \dfrac{ -\frac19 } { -\frac16 } = \frac23 \\ \end{array}\)
Sorry, this is not a geometric sequence.
+26375 Find the 2015th term of the geometric sequence 1,−1/3,-1/6,−1/9 ,.
In a geometric sequence, the common ratio, r between any two consecutive terms is always the same.
\(\begin{array}{rcl} \dfrac{ \text{Any term} } { \text{Previous term}} &=& \dfrac{u_n}{ u_n-1} = \text{ constant } = r \\ \end{array}\)
\(\begin{array}{rcr} 1,-\frac13,-\frac16,-\frac19,-\cdots \\ u_1 &=& 1\\ u_2 &=& -\frac13\\ u_3 &=& -\frac16\\ u_4 &=& -\frac19\\ \cdots \\ \end{array} \)
If three terms, \(u_n,\ u_{n+1},\ u_{n+2}\) are in geometric sequence, then:
\(\begin{array}{rcl} \dfrac{ u_{n+2} } { u_{n+1} } &=& \dfrac{ u_{n+1} } { u_{n} }\\ \end{array}\)
We have
\(\begin{array}{rcr} \dfrac{ u_2 } { u_1 } = \dfrac{ -\frac13 } { 1 } = -\frac13 &\ne& \dfrac{ u_3 } { u_2 } = \dfrac{ -\frac16 } { -\frac13 } = \frac12 \\ &\ne& \dfrac{ u_4 } { u_3 } = \dfrac{ -\frac19 } { -\frac16 } = \frac23 \\ \end{array}\)
Sorry, this is not a geometric sequence.
