geometric series help!

[ms.cupcake]

the sum of T3,T4,T5 is 1.75
the sum of T6,T7 is 6
find the first term,common ratio and n which the sum is less than 2100.

help me! anyone know how it work? the answer for first term is 1/6,common ratio=2,n=5.

 

[stapel]

anyone know how it work?

Did your class and textbook not cover this topic?

the sum of T3,T4,T5 is 1.75
the sum of T6,T7 is 6
find the first term,common ratio and n which the sum is less than 2100.

Do the variables stand for terms in the sequence or for partial sums of the series? (We can't see your book, so you need to tell us the definitions of the variables.)

 

[HallsofIvy]

the sum of T3,T4,T5 is 1.75
the sum of T6,T7 is 6
find the first term,common ratio and n which the sum is less than 2100.

help me! anyone know how it work? the answer for first term is 1/6,common ratio=2,n=5.

A "geometric series" is of the for a+ ar+ ar^2+ ar^3+ ... I assume that "T3" is the third term in the series, etc. (You really should say that in your post- this notation is not standard) then T3+ T4+ T5= ar^2+ ar^3+ ar^4= ar^2(1+ r+ r^2)= 1.75 and T6+ T7= ar^5+ ar^6= ar^5(1+ r)= 6. You have two equations to solve for a and r. Dividing one equation by the other will simpllify a lot.

 

[soroban]

Hello, ms.cupcake!

Even with excellent lectures, this problem is very difficult.

\(\displaystyle \text{The sum of }T_3,\,T_4,\,T_5\text{ is }\frac{7}{4}\)
\(\displaystyle \text{The sum of }T_6,\,T_7\text{ is }6.\)
\(\displaystyle \text{Find the first term, common ratio and }n\text{ for which the sum is less than 2100.}\)

We have: .\(\displaystyle T_3 = ar^2,\;T_4 = ar^3,\;T_5 = ar^4,\;T_6 = ar^5, \;T_7 = ar^6\)

We are told: .\(\displaystyle \begin{Bmatrix}ar^2 + ar^3 + ar^4 \:=\:\frac{7}{4} & \Rightarrow & ar^2(1+r+r^2) \:=\:\frac{7}{4} & [1] \\ ar^5 + ar^6 \:=\: 6 & \Rightarrow & ar^5(1+r) \:=\:6 & [2] \end{Bmatrix}\)

Divide [2] by [1]: .\(\displaystyle \dfrac{ar^5(1+r)}{ar^2(1+r+r^2)} \:=\:\dfrac{6}{\frac{7}{4}} \quad\Rightarrow\quad \dfrac{r^3(1+r)}{1+r+r^2} \:=\:\dfrac{24}{7}\)

This simplifies to: .\(\displaystyle 7r^4 + 7r^3 - 24r^2 - 24r - 24 \:=\:0\)

Now just solve the quartic equation . . .