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Tigertigre2000

05-31-2007, 11:22 AM

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.

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

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

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

\L a =5

Tigertigre2000

05-31-2007, 11:39 AM

...... How did u get 121? :?

skeeter

05-31-2007, 11:54 AM

121 = 3<sup>0</sup> + 3<sup>1</sup> + 3<sup>2</sup> + 3<sup>3</sup> + 3<sup>4</sup>

soroban

05-31-2007, 12:02 PM

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"?

Have you had Geometric Series?

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

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

The sum of the first n terms is: \: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: \:605\:=\:a\cdot\frac{1\,-\,3^5}{1\,-\,5}\:=\:a\cdot\frac{-242}{-2}

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

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