Consider an A.P.: a₁, a₂, … , aₙ; a₁ > 0. If a₂ − a₁ = −3/4, aₙ = (1/4)a₁, and ∑ᵢ₌₁ⁿ aᵢ = 525/2, then ∑ᵢ₌₁¹⁷ aᵢ is equal to

Consider an A.P.: a₁, a₂, … , aₙ; a₁ > 0. If a₂ − a₁ = −3/4, aₙ = (1/4)a₁, and Σ aᵢ = 525/2, then Σ aᵢ up to 17 terms is

Q. Consider an A.P.: $a_1, a_2, \ldots, a_n; \; a_1>0.$ If $$ a_2-a_1=-\frac{3}{4}, \quad a_n=\frac14 a_1, $$ and $$ \sum_{i=1}^n a_i=\frac{525}{2}, $$ then $$ \sum_{i=1}^{17} a_i $$ is equal to

(A) 238

(B) 136

(D) 952

Correct Answer: 238

Explanation

Given $a_2-a_1=d$, so

$$ d=-\frac34 $$

General term:

$$ a_n=a_1+(n-1)d $$

Using $a_n=\frac14 a_1$:

$$ a_1+(n-1)\left(-\frac34\right)=\frac14 a_1 $$

$$ a_1-\frac34(n-1)=\frac14 a_1 $$

$$ \frac34 a_1=\frac34(n-1) $$

$$ a_1=n-1 $$

Sum of $n$ terms:

$$ S_n=\frac{n}{2}\left[2a_1+(n-1)d\right] $$

Substitute $a_1=n-1$ and $d=-\frac34$:

$$ \frac{n}{2}\left[2(n-1)-\frac34(n-1)\right]=\frac{525}{2} $$

$$ \frac{n}{2}\cdot\frac54(n-1)=\frac{525}{2} $$

$$ \frac{5n(n-1)}{8}=\frac{525}{2} $$

$$ 5n(n-1)=2100 $$

$$ n(n-1)=420 $$

$$ n=21 $$

Hence,

$$ a_1=n-1=20 $$

Now sum of first 17 terms:

$$ S_{17}=\frac{17}{2}\left[2(20)+16\left(-\frac34\right)\right] $$

$$ =\frac{17}{2}(40-12) $$

$$ =\frac{17}{2}\times28=238 $$

Explanation

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