KCET · Maths · Sequences and Series
If the middle term of the AP is 300 , then the sum of its first 51 terms is
- A 15300
- B 14800
- C 16500
- D 14300
Answer & Solution
Correct Answer
(A) 15300
Step-by-step Solution
Detailed explanation
Given, number of terms, \(n=51\)
\(\because n\) is odd.
\(\therefore\) Middle term will be \(\left(\frac{n+1}{2}\right)\) th term.
\(\begin{aligned}
&=\left(\frac{51+1}{2}\right) \text { th term }=26 \text { th term } \\
\therefore \quad & T_{26}=300
\end{aligned}\)
\(a+25 d=300 \quad\left[\because T_{n}=a+(n-1) d\right]\)
\(\Rightarrow \quad T_{1}+25 d=300 \quad\left[\because T_{1}=a\right]\)
\(T_{1}=300-25 d\)
and \(T_{51}=a+50 d=T_{1}+50 d=300-25 d+50 d\)
\(=300+25 d\)
\(\therefore S_{51}=\frac{51}{2}[300-25 d+300+25 d]\)
\(\quad=\frac{51}{2}[600]=15300\)
\(\because n\) is odd.
\(\therefore\) Middle term will be \(\left(\frac{n+1}{2}\right)\) th term.
\(\begin{aligned}
&=\left(\frac{51+1}{2}\right) \text { th term }=26 \text { th term } \\
\therefore \quad & T_{26}=300
\end{aligned}\)
\(a+25 d=300 \quad\left[\because T_{n}=a+(n-1) d\right]\)
\(\Rightarrow \quad T_{1}+25 d=300 \quad\left[\because T_{1}=a\right]\)
\(T_{1}=300-25 d\)
and \(T_{51}=a+50 d=T_{1}+50 d=300-25 d+50 d\)
\(=300+25 d\)
\(\therefore S_{51}=\frac{51}{2}[300-25 d+300+25 d]\)
\(\quad=\frac{51}{2}[600]=15300\)
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