YP=+8+10+12 + ... + 106 and Q=2+4+6+8 + ... 104 are sums of arithmetic sequences, determine which is greater, Por Q, and by how much

Question

Kiana Sawyer

Mathematics

11 February, 18:11

YP=+8+10+12 + ... + 106 and Q=2+4+6+8 + ... 104 are sums of arithmetic sequences, determine which is greater, Por Q, and by how much

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Answers (1)

Know the Answer?

Clarissa Holmes · Answer 1

P is greater by 94.

Step-by-step explanation:

P = 8 + 10 + 12 + ... + 102 + 104 + 106

Q = 2 + 4 + 6 + 8 + 10 + 12 + ... + 102 + 104

The terms from 8 + 10 + ... + 104 are the same on both P and Q.

Therefore, the different terms are the 106 in P and the 2 + 4 + 6 in Q.

Clearly 106 > 2 + 4 + 6, so P is greater.

To know by how much we would subtract 2 + 4 + 6 = 12 from 106.

106 - 12 = 94.