Question ID: #865
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
- (1) 476
- (2) 952
- (3) 238
- (4) 136
Solution:
Given common difference $d = a_2 – a_1 = -\frac{3}{4}$.
Given last term $a_n = \frac{a_1}{4}$.
Given sum $S_n = \frac{525}{2}$.
Formula for $a_n$:
$$a_n = a_1 + (n-1)d$$
$$\frac{a_1}{4} = a_1 + (n-1)\left(-\frac{3}{4}\right)$$
$$\frac{a_1}{4} – a_1 = -\frac{3}{4}(n-1)$$
$$-\frac{3a_1}{4} = -\frac{3}{4}(n-1)$$
$$a_1 = n-1 \Rightarrow n = a_1 + 1$$
Formula for Sum $S_n$:
$$S_n = \frac{n}{2}(a_1 + a_n)$$
$$\frac{525}{2} = \frac{n}{2}\left(a_1 + \frac{a_1}{4}\right)$$
$$525 = n \left(\frac{5a_1}{4}\right)$$
$$105 = \frac{n a_1}{4}$$
$$n a_1 = 420$$
Substitute $n = a_1 + 1$:
$$(a_1 + 1)a_1 = 420$$
$$a_1^2 + a_1 – 420 = 0$$
$$(a_1 + 21)(a_1 – 20) = 0$$
Since $a_1 > 0$, we get $a_1 = 20$.
Then $n = 21$.
We need to find sum of first 17 terms:
$$S_{17} = \frac{17}{2}[2a_1 + (17-1)d]$$
$$S_{17} = \frac{17}{2}[2(20) + 16(-\frac{3}{4})]$$
$$S_{17} = \frac{17}{2}[40 – 12]$$
$$S_{17} = \frac{17}{2}[28] = 17 \times 14$$
$$S_{17} = 238$$
Ans. (3)
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