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5 + 8 + 11 + ... + 53

 

here's what i did:

\(S_{53} = {53 [2(5) + (53 - 1)3] \over 2}\)

\(S_{53} = {53 (166) \over 2}\)

\(S_{53} = 4399\)

 

the answer is 493, what did i do wrong?

 Nov 1, 2017

Best Answer 

 #2
avatar
+1

N =((L - F) / D) + 1, where N=Number of terms, L=Last term, F=First term, D=Common difference

N =((53 - 5) / 3) + 1

N = 17 Number of terms.

 

S =(F+L) / 2  x  N  , where F=First term, L=Last term. N=Number of terms, S=The Sum.

S =(53+5)/2  x  17 

S =58/2  x  17 

S =29   x  17 =493

 Nov 1, 2017
 #1
avatar+98197 
+1

 

 

Let's arrange the series in a slightly different manner

 

53 +  [ 50 + 5 ]  +  [ 47 + 8 ] +  [ 44 + 11]  , etc.

 

So....we have some "n"  pairs each summing to 55  plus the last term

 

Finding the number of pairs is a little tricky....but we can use this "formula"

 

n  =  (  [ last even term - first term ] / 3   + 1 )  / 2

 

So....the number of pairs is

 

 ([ 50 - 5 ] / 3  +   1)  / 2  =     

 

(45/ 3  + 1) / 2  =

 

[15 + 1 ] / 2 =

 

16 / 2  =

 

8

 

So......the sum is``    53  + (  8 * 55 )  =    493

 

 

 

cool cool cool 

 Nov 1, 2017
 #2
avatar
+1
Best Answer

N =((L - F) / D) + 1, where N=Number of terms, L=Last term, F=First term, D=Common difference

N =((53 - 5) / 3) + 1

N = 17 Number of terms.

 

S =(F+L) / 2  x  N  , where F=First term, L=Last term. N=Number of terms, S=The Sum.

S =(53+5)/2  x  17 

S =58/2  x  17 

S =29   x  17 =493

Guest Nov 1, 2017
 #3
avatar+21869 
+1

determining the sum of this arithmetic series?

5 + 8 + 11 + ... + 53

 

arithmetic series:

\(\begin{array}{lrrrrrrrrrrrrrrrrr} & {\color{red}d_0 = 5} && 8 && 11 && 14 && 17 && \cdots && 53 \\ \text{1. Difference } && {\color{red}d_1 = 3} && 3 && 3 && 3 && \cdots && 3 \\ \end{array}\)

 

Formula:

\(\begin{array}{|rcl|} \hline a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } \\ s &=& \binom{n}{1}\cdot {\color{red}d_0 } + \binom{n}{2}\cdot {\color{red}d_1 } \\ \hline \end{array}\)

 

n = ?

\(\begin{array}{|rcl|} \hline a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } \quad & | \quad a_n = 53 \quad {\color{red}d_0 } = 5 \quad {\color{red}d_1 } = 3 \\\\ 53 &=& \binom{n-1}{0}\cdot {\color{red}5 } + \binom{n-1}{1}\cdot {\color{red}3 } \\ 53 &=& 5+(n-1)\cdot 3 \\ 53 &=& 2 +3n \quad & | \quad -2\\ 51 &=& 3n \quad & | \quad :3 \\ 17 &=& n \\ \mathbf{n} &\mathbf{=}& \mathbf{17} \\ \hline \end{array} \)

 

sum (s) = ?

\(\begin{array}{|rcl|} \hline s &=& \binom{n}{1}\cdot {\color{red}d_0 } + \binom{n}{2}\cdot {\color{red}d_1 } \quad & | \quad n = 17 \quad {\color{red}d_0 } = 5 \quad {\color{red}d_1 } = 3 \\\\ s &=& \binom{17}{1}\cdot {\color{red}5 } + \binom{17}{2}\cdot {\color{red}3 } \\ s &=& 17\cdot 5 + \frac{17}{2} \cdot \frac{16}{1}\cdot 3 \\ s &=& 17\cdot \left(5+\frac{16}{2}\cdot 3 \right) \\ s &=& 17\cdot(5+ 24) \\ s &=& 17\cdot 29 \\ \mathbf{s} &\mathbf{=}& \mathbf{493} \\ \hline \end{array}\)

 

The sum is 493

 

laugh

 Nov 1, 2017

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