Let a_n be a sequence such that a_1 = 5 and a | vedclass

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(D) Given the recurrence relation $a_{n+1} - a_n = n - 2$.We can write this for values from $n = 1$ to $n = 50$:$a_2 - a_1 = 1 - 2 = -1$$a_3 - a_2 = 2

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$1180$

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(D) Given the recurrence relation $a_{n+1} - a_n = n - 2$.We can write this for values from $n = 1$ to $n = 50$:$a_2 - a_1 = 1 - 2 = -1$$a_3 - a_2 = 2 - 2 = 0$$a_4 - a_3 = 3 - 2 = 1$...$a_{51} - a_{50} = 50 - 2 = 48$Summing these equations,the intermediate terms cancel out:$a_{51} - a_1 = (-1) + 0 + 1 + 2 + \dots + 48$The sum on the right is an arithmetic progression with $n = 50$ terms,first term $a = -1$,and last term $l = 48$.Sum $S = \frac{n}{2}(a + l) = \frac{50}{2}(-1 + 48) = 25 	imes 47 = 1175$.Thus,$a_{51} - 5 = 1175$,which gives $a_{51} = 1180$.

Let $a_n$ be a sequence such that $a_1 = 5$ and $a_{n+1} = a_n + (n - 2)$ for all $n \in N$. Then $a_{51}$ is:

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