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The $12 t h$ term from the end of the AP, –2, –4, –6, ..., –100 is[[1]].

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Detailed Solution

The

12 t h

term from the end of the AP, –2, –4, –6, ..., –100 is -78.
The given AP series is –2, –4, –6, ..., –100.
As we reverse the given series,
Last term becomes the first term so,

a = − 100

.
The common difference

d

is,

d = − 4 − (− 6) = − 4 + 6 = 2

Then, the 12th term from last will come from this equation,

a n = a + n − 1 d

a 12 = − 100 + 12 − 1 2 = − 100 + 112 = − 100 + 22 = − 78

Thus, the 12th term from last is -78.

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