Question

If the sum of first 75 terms of an A.P. is 2625, then the \( 38^{th} \) term of the A.P. is:

• A. \( 39 \)
• B. \( 37 \)
• C. \( 36 \)
• D. \( 38 \)
• E. \( 35 \)

Hint:

If you find division by 75 hard, remember that \( 75 \times 4 = 300 \). Since \( 300 \times 8 = 2400 \), you know the answer is around \( 4 \times 8 = 32 \). Adding three more 75s (\(225\)) gets you exactly to 2625, so \( 32 + 3 = 35 \).

Answer 1

Answer: (E)

Concept: For an A.P. with an odd number of terms \( n \), the sum \( S_n \) is related to the middle term by: \[ S_n = n \times (\text{Middle Term}) \] The middle term of \( n \) terms is the \( \left(\frac{n+1}{2}\right)^{th} \) term.

Step 1: Identify the middle term.
For \( n = 75 \), the middle term is: \[ \text{Middle Term} = \frac{75 + 1}{2} = 38^{th} \text{ term} \]

Step 2: Set up the equation using the sum property.
\[ S_{75} = 75 \times a_{38} \] Given \( S_{75} = 2625 \).

Step 3: Solve for the \( 38^{th} \) term.
\[ 2625 = 75 \times a_{38} \] \[ a_{38} = \frac{2625}{75} \] Calculation: \( 2625 / 75 = 35 \).