Sequences

T

Tigertigre2000

New member
Joined
Aug 20, 2006
Messages
23
a, 3a

The first tem in the sequence above is "a",
and each term after the first is 3 times the preceding term.
If the sum of the first 5 terms is 605, what is the value of "a"?


How do u solve it.... is there a certain formula for this problem?
 
It is just a regular algebra problem in disguise.

\(\displaystyle \L \sum_{k=0}^4 3^ka \,\, = \,\, 605\)

\(\displaystyle \L a\sum_{k=0}^4 3^k \,\, = \,\, 605\)

\(\displaystyle \L a(121) \,\, = \,\, 605\)

\(\displaystyle \L a =5\)
 
121 = 3<sup>0</sup> + 3<sup>1</sup> + 3<sup>2</sup> + 3<sup>3</sup> + 3<sup>4</sup>
 
Hello, Tigertigre2000!

a, 3a

The first term in the sequence above is "a",
and each term after the first is 3 times the preceding term.
If the sum of the first 5 terms is 605, what is the value of "a"?
Click to expand...

Have you had Geometric Series?

This is a geometric series with first term \(\displaystyle a\) and common ratio \(\displaystyle r\,=\,3\).

The \(\displaystyle n^{th}\) term is: \(\displaystyle \:a_n \:=\:a\cdot r^{n-1}\)

The sum of the first \(\displaystyle n\) terms is: \(\displaystyle \:S_n \:=\:a\cdot\frac{1\,-\,r^n}{1\,-\,r}\)


We are told that the sum of the first 5 terms is 605.

So we have: \(\displaystyle \:605\:=\:a\cdot\frac{1\,-\,3^5}{1\,-\,5}\:=\:a\cdot\frac{-242}{-2}\)

Therefore: \(\displaystyle \:121a \:=\:605\;\;\Rightarrow\;\;\fbox{a\:=\:5}\)

 
You must log in or register to reply here.
Top