Mathematics

Mathematics

Problem

For an arithmetic progression with first term aa, common difference dd, and an odd number of terms nn, a student writes Sn=n−12[2a+(n−1)d]+(a+n−12d).S_n=\frac{n-1}{2}\bigl[2a+(n-1)d\bigr]+\Bigl(a+\frac{n-1}{2}d\Bigr). Which of the following equalities allows this expression to be simplified to the standard form Sn=n2[2a+(n−1)d]\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]? a) a+n−12d=1n[2a+(n−1)d]a+\dfrac{n-1}{2}d=\dfrac{1}{n}[2a+(n-1)d] b) a+n−12d=1n−1[2a+(n−1)d]a+\dfrac{n-1}{2}d=\dfrac{1}{n-1}[2a+(n-1)d] c) a+n−12d=12[2a+(n−1)d]a+\dfrac{n-1}{2}d=\dfrac{1}{2}[2a+(n-1)d] d) a+n−12d=[2a+(n−1)d]a+\dfrac{n-1}{2}d=[2a+(n-1)d]

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