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88
2
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Find the value of $a_2+a_4+a_6+a_8+\dots+a_{98}$ if $a_1, a_2, a_3, \ldots$ is an arithmetic progression with common difference $1$ and\[a_1+a_2+a_3+\dots+a_{98}=137.\]

 Jan 6, 2021
 #1
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Since we are taking exactly haf the terms, a_2 + a_4 + ... + a_98 = 137/2.

 Jan 6, 2021
 #2
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That's not correct because each term is greater than the one before it. Since there are 49 pairs, we have to subtract 49 to get 88 divide by 2 to get 44 and add the 49 back. 93.

:D

 Jan 6, 2021

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