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(C) Given $S_{10} = 390$. Using the formula $S_n = \frac{n}{2}[2a + (n-1)d]$,we have:$\frac{10}{2}[2a + 9d] = 390 \Rightarrow 2a + 9d = 78$ $......(1)

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$790$

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(C) Given $S_{10} = 390$. Using the formula $S_n = \frac{n}{2}[2a + (n-1)d]$,we have:$\frac{10}{2}[2a + 9d] = 390 \Rightarrow 2a + 9d = 78$ $......(1)$The ratio of the tenth term $(t_{10})$ to the fifth term $(t_5)$ is $15:7$:$\frac{a + 9d}{a + 4d} = \frac{15}{7}$ $\Rightarrow 7(a + 9d) = 15(a + 4d)$ $\Rightarrow 7a + 63d = 15a + 60d$ $\Rightarrow 8a = 3d$ $\Rightarrow d = \frac{8a}{3}$ $......(2)$Substituting $(2)$ into $(1)$:$2a + 9(\frac{8a}{3}) = 78$ $\Rightarrow 2a + 24a = 78$ $\Rightarrow 26a = 78$ $\Rightarrow a = 3$Using $a = 3$ in $(2)$,$d = \frac{8(3)}{3} = 8$.We need to find $S_{15} - S_5$:$S_{15} - S_5 = \frac{15}{2}[2(3) + 14(8)] - \frac{5}{2}[2(3) + 4(8)]$$= \frac{15}{2}[6 + 112] - \frac{5}{2}[6 + 32]$$= \frac{15}{2}(118) - \frac{5}{2}(38) = 15(59) - 5(19) = 885 - 95 = 790$.

Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10} = 390$ and the ratio of the tenth and the fifth terms is $15:7$,then $S_{15} - S_5$ is equal to:

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