Question

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

• A. 136
• B. 476
• C. 238
• D. 952

Hint:

Always express the last term of an A.P. using $a_n=a_1+(n-1)d$ to relate $a_1$ and $n$.

Answer 1

The Correct Option is A

Step 1: Identifying A.P. parameters.
Given \[ d=a_2-a_1=-\frac{3}{4} \] \[ a_n=a_1+(n-1)d=\frac{a_1}{4} \] \[ a_1+(n-1)\left(-\frac{3}{4}\right)=\frac{a_1}{4} \] \[ \Rightarrow \frac{3a_1}{4}=\frac{3(n-1)}{4} \Rightarrow a_1=n-1 \] Step 2: Using sum of $n$ terms.
\[ S_n=\frac{n}{2}(a_1+a_n)=\frac{525}{2} \] \[ \Rightarrow n\left(a_1+\frac{a_1}{4}\right)=525 \] \[ \Rightarrow \frac{5n a_1}{4}=525 \Rightarrow n a_1=420 \] Substitute $a_1=n-1$, \[ n(n-1)=420 \Rightarrow n=21 \] Thus, \[ a_1=20 \] Step 3: Calculating $\sum_{i=1^{17} a_i$.}
\[ S_{17}=\frac{17}{2}\left[2a_1+(17-1)d\right] \] \[ =\frac{17}{2}\left[40-12\right] =\frac{17}{2}\times 28 =136 \]