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If P = 7+9+ 11 +13 + ... + 101 and Q = 3 + 5 + 7 + 9 + ... + 99 are sums of arithmetic sequences, determine which is greater, Por Q, and by how much.

Sagot :

Answer:

The sum of arithmetic sequence is:

[tex]s = \frac{n}{2} (2a(1) + (n - 1)d)[/tex]

The general term of arithmetic sequence:

[tex]a(n) = a(1) + (n - 1)d[/tex]

***********************************For P:********************************************

[tex]a(1) = 7 \\ d = 2 \\ a(n) = 101\\ \\ a(n) = 7 + (n - 1)2 \\ a(n) = 7 + 2n - 2 \\ a(n) = 2n + 5 \\ 101 = 2n + 5 \\ 2n = 101 - 5 \\ 2n = 96 \\ \frac{2n}{2} = \frac{96}{2} \\ n = 48[/tex]

Now we can find the sum of the arithmetic sequence by substituting (n=48 and a(1)=7)

[tex]s = \frac{n}{2} (2a(1) + (n - 1)d) \\ s = \frac{48}{2} (2(7) + (48 - 1)2) \\ s = 24(14 + 94) \\ s = 24(108) \\ s = 2592[/tex]

************************************For Q:********************************************

[tex]a(1) = 3 \\ d = 2 \\ a(n) = 99 \\ a(n) = 3 + (n - 1)2 \\ 99 = 3 + 2n - 2 \\ 99 = 2n + 1 \\ 2n = 98 \\ n = 49[/tex]

Now we can find the sum of the arithmetic sequence by substituting (n=49 and a(1)=)

[tex]s = \frac{n}{2} (2a(1) + (n - 1)d) \\ s = \frac{49}{2} (2(3) + (49 - 1)2) \\ s = \frac{49}{2} (6 + 96) \\ s = \frac{49}{2} (102) \\ s = 2499[/tex]

S(P)= 2592

S(Q)= 2499

Sum of (P) =Sum of (Q) + 93

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