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Tigertigre2000

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Aug 20, 2006
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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 a\displaystyle a and common ratio r = 3\displaystyle r\,=\,3.

The nth\displaystyle n^{th} term is:  an = a⋅rn−1\displaystyle \:a_n \:=\:a\cdot r^{n-1}

The sum of the first n\displaystyle n terms is:  Sn = a⋅1 − rn1 − r\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:  605 = a⋅1 − 351 − 5 = a⋅−242−2\displaystyle \:605\:=\:a\cdot\frac{1\,-\,3^5}{1\,-\,5}\:=\:a\cdot\frac{-242}{-2}

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

 
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