AP: Common Difference
Let the n^{th} term of AP be defined as t_{n}, and sum up to 'n' terms be defined as S_{n}. If t_{8} = t_{16} and t_{3} is not equal to t_{7}, what is S_{23}?
 23(t_{16}  t_{8})
 0
 23t_{11}
 Cannot be determined

Correct Answer
Choice B. 0
Explanatory Answer
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Detailed Solution
t_{8}=t_{16}. This can happen under two scenarios t_{8} = t_{16} or t_{8} = – t_{16}.
If t_{8} = t_{16}, the common difference would be 0 suggesting that t_{3} would be equal to t_{7}. However, we know t_{3} is not equal to t_{7}, so the common difference cannot be zero.
This tells us that t_{8} = – t_{16} Or, t_{8} + t_{16} = 0.
If t_{8} + t_{16} = 0, then t_{12} = 0.
t_{12} = t_{8} + 4d, and t_{16} – 4d So, t_{12} =
.
For any two terms in an AP, the mean is the term right in between them. So, t
_{12} is the arithmetic mean of t
_{8} and t
_{16}.
So, t
_{12} = 0.
Now, S
_{23} = 23 × t
_{12}.
We know that average of n terms in an A.P. is the middle term. This implies that sum of n terms in an A.P., is n times the middle term.
So, S
_{23} = 0.
Correct Answer: zero.
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