Consider an A.P.: a_1, a_2, dots, a_n,with a_ | vedclass

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(C) Given the sum of an $A$.$P$. is $S_n = \frac{n}{2}(a_1 + a_n) = \frac{525}{2}$ and common difference $d = a_2 - a_1 = -\frac{3}{4}$.Substituting $

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$238$

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(C) Given the sum of an $A$.$P$. is $S_n = \frac{n}{2}(a_1 + a_n) = \frac{525}{2}$ and common difference $d = a_2 - a_1 = -\frac{3}{4}$.Substituting $a_n = \frac{1}{4} a_1$ into the sum formula:$\frac{n}{2}(a_1 + \frac{a_1}{4}) = \frac{525}{2} \implies \frac{n}{2}(\frac{5a_1}{4}) = \frac{525}{2} \implies \frac{5a_1 n}{8} = \frac{525}{2} \implies a_1 n = 420$.Using the formula $a_n = a_1 + (n-1)d$:$\frac{1}{4} a_1 = a_1 + (n-1)(-\frac{3}{4}) \implies -\frac{3}{4} a_1 = -\frac{3}{4}(n-1) \implies a_1 = n-1$.Substituting $a_1 = n-1$ into $a_1 n = 420$:$(n-1)n = 420 \implies n^2 - n - 420 = 0 \implies (n-21)(n+20) = 0$.Since $n > 0$,we have $n = 21$ and $a_1 = 21 - 1 = 20$.Now,calculate $\sum_{i=1}^{17} a_i = \frac{17}{2}[2a_1 + (17-1)d]$:$= \frac{17}{2}[2(20) + 16(-\frac{3}{4})] = \frac{17}{2}[40 - 12] = \frac{17}{2}[28] = 17 	imes 14 = 238$.

Consider an $A$.$P$.: $a_1, a_2, \dots, a_n$,with $a_1 > 0$. If $a_2 - a_1 = -\frac{3}{4}$,$a_n = \frac{1}{4} a_1$,and $\sum_{i=1}^n a_i = \frac{525}{2}$,then $\sum_{i=1}^{17} a_i$ is equal to:

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