Question

If the middle term of the A.P is 300 then the sum of its first 51 terms is

• A. 15300
• B. 14800
• C. 16500
• D. 14300

Answer 1

The Correct Option is A

The sum of the first \( n \) terms of an arithmetic progression (A.P.) is given by the formula: \[ S_n = \frac{n}{2} \times \left(2a + (n-1)d \right) \] where:
\( a \) is the first term,
\( d \) is the common difference,
\( n \) is the number of terms.
In this case, the number of terms is 51, and the middle term is given as 300. Since the middle term of an A.P. is the \( \left(\frac{51+1}{2}\right) = 26 \)-th term, we have: \[ a + (26-1)d = 300 \] Simplifying, we get: \[ a + 25d = 300 \] Now, using the formula for the sum of the first 51 terms: \[ S_{51} = \frac{51}{2} \times \left(2a + (51-1)d \right) \] \[ S_{51} = \frac{51}{2} \times \left(2a + 50d \right) \] Substituting \( a + 25d = 300 \) into the equation: \[ S_{51} = 51 \times 300 = 15300 \] Therefore, the sum of the first 51 terms is \( 15300 \).

Answer 2

Given that the middle term of an A.P. is 300 and we need to find the sum of its first 51 terms.

In an A.P., the middle term of an A.P. with an odd number of terms is the average of all the terms. Also, the sum of \(n\) terms of an A.P. is given by \(S_n = \frac{n}{2} [2a + (n-1)d]\), where \(a\) is the first term and \(d\) is the common difference.

Alternatively, \(S_n = \frac{n}{2} [a + l]\), where \(l\) is the last term.

Since we have 51 terms, the middle term is the \(\frac{51+1}{2} = 26\)th term. So, the 26th term is given to be 300.

Let the first term be \(a\) and the common difference be \(d\). Then the 26th term is \(a + 25d = 300\).

The sum of the first 51 terms is given by:

\(S_{51} = \frac{51}{2} [2a + (51-1)d] = \frac{51}{2} [2a + 50d] = 51[a + 25d]\).

We know that \(a + 25d = 300\), so we can substitute this value into the sum formula:

\(S_{51} = 51 \times 300 = 15300\).

Therefore, the sum of the first 51 terms is 15300.