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(A) The sum of the first $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.Given $S_{20} = \frac{20}{2}[2a + 19d] =

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$395$

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(A) The sum of the first $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.Given $S_{20} = \frac{20}{2}[2a + 19d] = 790$,which simplifies to $2a + 19d = 79$ $(1)$.Given $S_{10} = \frac{10}{2}[2a + 9d] = 145$,which simplifies to $2a + 9d = 29$ $(2)$.Subtracting $(2)$ from $(1)$,we get $(2a + 19d) - (2a + 9d) = 79 - 29$,so $10d = 50$,which means $d = 5$.Substituting $d = 5$ into $(2)$,we get $2a + 9(5) = 29$,so $2a + 45 = 29$,which gives $2a = -16$,or $a = -8$.We need to find $S_{15} - S_5 = \frac{15}{2}[2a + 14d] - \frac{5}{2}[2a + 4d]$.Substituting $a = -8$ and $d = 5$:$S_{15} - S_5 = \frac{15}{2}[2(-8) + 14(5)] - \frac{5}{2}[2(-8) + 4(5)]$$= \frac{15}{2}[-16 + 70] - \frac{5}{2}[-16 + 20]$$= \frac{15}{2}(54) - \frac{5}{2}(4)$$= 15(27) - 5(2) = 405 - 10 = 395$.

Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{20} = 790$ and $S_{10} = 145$,then $S_{15} - S_5$ is:

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