Question

The n term of a geometric sequence is denoted by Tn and the sum of the first n terms is denoted by Sn.Given T6-T4=5/2 and S5-S3=5.Calculate (a)the common ratio. (b)the first term of this geometric sequence

Answer 1

1 step: $S_{5}=T_{1}+T_{2}+T_{3}+T_{4}+T_{5}$, $S_{3}=T_{1}+T_{2}+T_{3}$, then
 $S_{5}-S_{3}=T_{4}+T_{5}=5$.

2 step: $T_{n}=T_{1}*q^{n-1}$, then 
$T_{6}=T_{1}*q^{5}$
$T_{5}=T_{1}*q^{4}$
$T_{4}=T_{1}*q^{3}$
$T_{3}=T_{1}*q^{2}$
and $\left \{ {{T_{6}-T_{4}= \frac{5}{2} } \atop {T_{5}+T_{4}=5}} \right.$ will have form $\left \{ {{T_1*q^{5}-T_{1}*q^{3}= \frac{5}{2} } \atop {T_{1}*q^{4}+T_{1}*q^{3}=5} \right.$.

3 step: Solve this system  $\left \{ {{T_1*q^{3}*(q^{2}-1)= \frac{5}{2} } \atop {T_{1}*q^{3}*(q+1)=5} \right.$ and dividing first equation on second we obtain $\frac{q^{2}-1}{q+1}= \frac{ \frac{5}{2} }{5}$. So, $\frac{(q-1)(q+1)}{q+1} = \frac{1}{2}$ and $q-1= \frac{1}{2}$, $q= \frac{3}{2}$ - the common ratio.

4 step: Insert $q= \frac{3}{2}$into equation $T_{1}*q^{3}*(q+1)=5$ and obtain $T_{1}* \frac{27}{8}*( \frac{3}{2}+1 ) =5$, from where $T_{1}= \frac{16}{27}$.