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The 10th term of the AP $3, 7, 11, 15, \ldots$ is:
A$40$, mistakenly multiplying 10 by 4 (the difference)
B$39$, since $a_{10} = 3 + 9\cdot 4 = 39$ by $a + (n-1)d$
C$30$, taking 10 times the first term as the answer
D$43$, summing 3 and $10\cdot 4$ without the $-1$
Answer & Solution
Correct answer: B. $39$, since $a_{10} = 3 + 9\cdot 4 = 39$ by $a + (n-1)d$
$a_{10} = a + 9d = 3 + 36 = 39$.
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