Consider an A.P.: a 1 , a 2 , … , a n ; a 1 0 . If a 2 - a 1 = - 3 4 , a n = 1 4 a 1 and i = 1 n a i = 525 2 , then i = 1 17 a i is equal to: [2026]

Question

Consider an A.P.: $a_{1}, a_{2}, \dots, a_{n}$ ; $a_{1} > 0$ . If $a_{2} - a_{1} = \frac{- 3}{4}, a_{n} = \frac{1}{4} a_{1}$ and $\sum_{i = 1}^{n} a_{i} = \frac{525}{2},$ then $\sum_{i = 1}^{17} a_{i}$ is equal to: [2026]

Smart Achievers · Accepted Answer

(4)

$S_{n} = \frac{n}{2} (a_{1} + a_{n}) = \frac{525}{2}, d = \frac{- 3}{4}$

$\frac{n}{2} [a_{1} + \frac{a_{1}}{4}] = \frac{525}{2}$

$\frac{5 a_{1} n}{4} = 525$

$a_{1} n = 420$

$a_{n} = a_{1} + (n - 1) (\frac{- 3}{4})$

$\Rightarrow \frac{- 3}{4} a_{1} = (\frac{- 3}{4}) (n - 1) \Rightarrow a_{1} = n - 1$

$n (n - 1) = 420$

$n^{2} - n - 420 = 0$

$(n - 21) (n + 20) = 0$

$n = 21, a_{1} = 20$

$\sum_{i = 1}^{17} a_{i} = \frac{17}{2} [2 a_{1} + 16 d]$

$= \frac{17}{2} [40 + 16 (\frac{- 3}{4})]$

$= \frac{17}{2} [40 - 12]$

$= 17 \times 14 = 238$

Solution

Q.
Consider an A.P.: $a_{1}, a_{2}, \dots, a_{n}$ ; $a_{1} > 0$ . If $a_{2} - a_{1} = \frac{- 3}{4}, a_{n} = \frac{1}{4} a_{1}$ and $\sum_{i = 1}^{n} a_{i} = \frac{525}{2},$ then $\sum_{i = 1}^{17} a_{i}$ is equal to: [2026]

1 952

2 136

3 476

4 238

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