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**Punch** The sum of the first three terms and that of the last three terms of an arithmetic progression are 9 and 57 respectively. It is known that the sum of the whole sequence is 121, find the number of terms in this sequence.

$\displaystyle S_3=\frac{3}{2}[2a+(3-1)d]=9$ -----(1)

$\displaystyle S_n-S_{n-3}=121-\frac{n-3}{2}[2a+(n-3-1)d]=57$

$\displaystyle 121-\frac{n-3}{2}[2a+(n-4)d]=57$

$\displaystyle \frac{n-3}{2}[2a+(n-4)d]=64$ -----(2)

$\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]=121$ -------(3)