Verify that each of the following is an $AP$,and then write its next three terms.
$a+b, (a+1)+b, (a+1)+(b+1), \ldots$
$a+b, (a+1)+b, (a+1)+(b+1), \ldots$
(N/A) Given sequence: $a_1 = a+b, a_2 = a+1+b, a_3 = a+1+b+1$.
Calculate the common difference $d$:
$d_1 = a_2 - a_1 = (a+1+b) - (a+b) = 1$.
$d_2 = a_3 - a_2 = (a+1+b+1) - (a+1+b) = 1$.
Since $d_1 = d_2 = 1$,the common difference is constant,so the sequence is an $AP$.
The next three terms are:
$a_4 = a_3 + d = (a+1+b+1) + 1 = a+b+3 = (a+2)+(b+1)$.
$a_5 = a_4 + d = (a+b+3) + 1 = a+b+4 = (a+2)+(b+2)$.
$a_6 = a_5 + d = (a+b+4) + 1 = a+b+5 = (a+3)+(b+2)$.
Calculate the common difference $d$:
$d_1 = a_2 - a_1 = (a+1+b) - (a+b) = 1$.
$d_2 = a_3 - a_2 = (a+1+b+1) - (a+1+b) = 1$.
Since $d_1 = d_2 = 1$,the common difference is constant,so the sequence is an $AP$.
The next three terms are:
$a_4 = a_3 + d = (a+1+b+1) + 1 = a+b+3 = (a+2)+(b+1)$.
$a_5 = a_4 + d = (a+b+3) + 1 = a+b+4 = (a+2)+(b+2)$.
$a_6 = a_5 + d = (a+b+4) + 1 = a+b+5 = (a+3)+(b+2)$.
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| Column $A$ | Column $B$ |
| $(A_{1}) \quad 2, -2, -6, -10, \ldots$ | $(B_{1}) \quad \frac{2}{3}$ |
| $(A_{2}) \quad a = -18, n = 10, a_{n} = 0$ | $(B_{2}) \quad -5$ |
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| $(A_{4}) \quad a_{2} = 13, a_{4} = 3$ | $(B_{4}) \quad -4$ |
| $(B_{5}) \quad 2$ | |
| $(B_{6}) \quad \frac{1}{2}$ | |
| $(B_{7}) \quad 5$ |
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