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the answer is B
17=a1+2d
65=a1+18d
subtraction we get d=3
17=a1+2*3 so a1=11
a10=11+3(9)=38
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Bunuel
Let T be a sequence of the form \(a_n=a_1+d*(n-1)\). If \(a_3=17\) and \(a_{19}=65\), find \(a_{10}\).

A. 37
B. 38
C. 39
D. 40
E. 41

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MAGOOSH OFFICIAL SOLUTION:

The formula in the first sentence tells us that this is an arithmetic sequence. The first term and the common difference are unknown, but we can generate two equations from the values of the two terms given.
a3 = a1 + 2d = 17
a19 = a1 + 18d = 65

Subtract the first equation from the second, and we get 16d = 48, which means d = 3. From the value of third term, we can see that first term must equal 11. Therefore, a10 = 11 + 3*9 = 38.

Answer = (B).
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Test please ignore
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