Consider an A.P.: $a_{1},a_{2},\ldots.,a_{n};a_{1} 0$. If $a_{2} - a_{1}$$= \frac{- 3}{4},a_{n}

Question

Consider an A.P.: $a_{1},a_{2},\ldots.,a_{n};a_{1} > 0$. If $a_{2} - a_{1}$

$= \frac{- 3}{4},a_{n} = \frac{1}{4}a_{1}$, and

$\sum_{i = 1}^{n}\mspace{2mu} a_{i} = \frac{525}{2}$, then $\sum_{i = 1}^{17}\mspace{2mu} a_{i}$ is equal to

Accepted Answer

$S_{n} = \frac{n}{2}\left\lbrack a_{1} + a_{n} \right\rbrack = \frac{525}{2},\text{ }d = \frac{- 3}{4}$

$$\begin{matrix} & \frac{n}{2}\left\lbrack a_{1} + \frac{a_{1}}{4} \right\rbrack = \frac{525}{2} \\ & \frac{5a_{1}n}{4} = 525 \\ & a_{1}n = 420 \\ & a_{n} = a_{1} + (n - 1)\left( \frac{- 3}{4} \right) \\ & \ \Rightarrow \frac{- 3}{4}a_{1} = \left( \frac{- 3}{4} \right)(n - 1) \Rightarrow a_{1} = n - 1 \\ & n(n - 1) = 420 \\ & n^{2} - n - 420 = 0 \\ & \ (n - 21)(n + 20) = 0 \\ & n = 21,a_{1} = 20 \\ & \ \sum_{i = 1}^{17}\mspace{2mu}\mspace{2mu} a_{i} = \frac{17}{2}\left\lbrack 2a_{1} + 16d \right\rbrack \\ & \ = \frac{17}{2}\left\lbrack 40 + 16\left( \frac{- 3}{4} \right) \right\rbrack \\ & \ = \frac{17}{2}\lbrack 40 - 12\rbrack \\ & \ = 17 \times 14 = 238 \end{matrix}$$

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