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a_1, a_2, a_3, ..., a_{99} is an arithmetic progeression.  If a_2 + a_5 + a_8 + ... + a_{98} = 205, then find a_1 + a_2 + a_3 + ... + a_{99}.  Be sure to show all your work.

Dec 3, 2019

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$$a_1, a_2, a_3, \cdots , a_{99}$$ is an arithmetic progeression.
If $$a_2 + a_5 + a_8 + \cdots+ a_{98} = 205$$,
then find $$a_1 + a_2 + a_3 + \cdots + a_{99}$$.

Formula arithmetic progeression: $$a_n = a_1 + (n-1)d$$

$$\begin{array}{|lrcll|} \hline a_2 =& a_1 + (2-1)*d &=& a_1 + d \\ a_5 =& a_1 + (5-1)*d &=& a_1 + 4d \\ a_8 =& a_1 + (8-1)*d &=& a_1 + 7d \\ a_{11} =& a_1 + (11-1)*d &=& a_1 + 10d \\ \cdots \\ a_{98} =& a_1 + (98-1)*d &=& a_1 + 97d \\ \hline \text{terms:} & 2+3(i-1) = 98 \\ & 3(i-1) = 96 \\ & i-1 = 32 \\ & i = 33 \\ \hline \text{sum}& 205&=& 33*a_1+d(1+4+7+10+\cdots + 97 ) \\ & 205&=& 33*a_1+d\left(\dfrac{1+97}{2}\right)\times 33 \\ &\mathbf{ 205 }&=& \mathbf{ 33*a_1+49\cdot 33d } \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline a_1 =& a_1 &=& a_1 \\ a_2 =& a_1 + (2-1)*d &=& a_1 + d \\ a_3 =& a_1 + (3-1)*d &=& a_1 + 2d \\ a_4 =& a_1 + (4-1)*d &=& a_1 + 3d \\ a_5 =& a_1 + (5-1)*d &=& a_1 + 4d \\ \cdots \\ a_{99} =& a_1 + (99-1)*d &=& a_1 + 98d \\ \hline \text{terms:} & 99 \\ \hline & \text{sum} &=& 99*a_1+d(1+2+3+4+\cdots + 98 ) \\ & \text{sum}&=& 99*a_1+d\left(\dfrac{1+98}{2}\right)\times 98 \\ &\mathbf{ \text{sum} }&=& \mathbf{ 99*a_1+49\cdot 99d } \\ & \text{sum} &=& 3\times \left( \underbrace{33*a_1+49\cdot 33d}_{=205} \right) \\ & \text{sum} &=& 3\times 205 \\ & \mathbf{ \text{sum} } &=& \mathbf{615} \\ \hline \end{array}$$

$$a_1 + a_2 + a_3 + \cdots + a_{99} = 615$$

Dec 3, 2019