Consider an A.P.

Question

Consider an A.P. $a_1, a_2, \ldots, a_n$ with $a_1>0$. If $a_2-a_1 = -\dfrac34$, $a_n = \dfrac14 a_1$, and $\displaystyle\sum_{i=1}^{n} a_i = \dfrac{525}{2}$, then $\displaystyle\sum_{i=1}^{17} a_i$ is equal to:

Accepted Answer

Common difference $d = -\tfrac34$. Using $S_n = \dfrac{n}{2}(a_1+a_n) = \dfrac{n}{2}\left(a_1+\tfrac14a_1\right) = \dfrac{5a_1n}{8} = \dfrac{525}{2}$:
$$5a_1n = 2100 \implies a_1 n = 420. \quad (1)$$

Also $a_n = a_1+(n-1)d = \tfrac14 a_1 \Rightarrow (n-1)\left(-\tfrac34\right) = -\tfrac34 a_1 \Rightarrow n-1 = a_1 \Rightarrow n = a_1+1$.

Substituting into (1): $a_1(a_1+1) = 420 \Rightarrow a_1^2+a_1-420=0 \Rightarrow (a_1+21)(a_1-20)=0$. Since $a_1>0$, $a_1=20$.

$$\sum_{i=1}^{17} a_i = \frac{17}{2}\left[2(20)+16\left(-\frac34\right)\right] = \frac{17}{2}[40-12] = 17\times14 = 238.$$

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