Question

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

• A. 231
• B. 234
• C. 236
• D. 238

Hint:

In A.P. problems, always use the $n$th term relation first to connect $a$, $d$, and $n$ before applying the sum formula.

Answer 1

The Correct Option is D

Step 1: Identify the A.P. parameters.
First term: \[ a_1=a \] Common difference: \[ d=a_2-a_1=-\frac{3}{4} \] Last term: \[ a_n=\frac{a}{4} \] Step 2: Use the $n$th term formula. 
\[ a_n=a+(n-1)d \Rightarrow \frac{a}{4}=a-(n-1)\frac{3}{4} \] \[ \Rightarrow a=\;n-1 \] Step 3: Use the sum of $n$ terms formula. 
\[ \sum_{i=1}^{n} a_i=\frac{n}{2}(a_1+a_n) =\frac{n}{2}\left(a+\frac{a}{4}\right) =\frac{5an}{8} \] Given: \[ \frac{5an}{8}=\frac{525}{2} \Rightarrow an=420 \] Step 4: Substitute $a=n-1$. 
\[ n(n-1)=420 \Rightarrow n=21 \] \[ a=20 \] Step 5: Find $\sum_{i=1^{17} a_i$.} 
\[ S_{17}=\frac{17}{2}\left[2a+(17-1)d\right] \] \[ =\frac{17}{2}\left[40-12\right] =\frac{17}{2}\times28 =238 \]