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)= (5)/(2) } \atop {T_(5)+T_(4)=5}} \right.` will have form
`\left \{ {{T_1*q^(5)-T_(1)*q^(3)= (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)= (5)/(2) } \atop {T_(1)*q^(3)*(q+1)=5} \right.` and dividing first equation on second we obtain
`(q^(2)-1)/(q+1)= ( (5)/(2) )/(5)`. So,
`((q-1)(q+1))/(q+1) = (1)/(2)` and
`q-1= (1)/(2)`,
`q= (3)/(2)` - the common ratio.

4 step: Insert
`q= (3)/(2)`into equation
`T_(1)*q^(3)*(q+1)=5` and obtain
`T_(1)* (27)/(8)*( (3)/(2)+1 ) =5`, from where
`T_(1)= (16)/(27)`.