Question

If the sum of 12th and 22nd terms of an AP is 100, then the sum of the first 33 terms of an AP is

• A. 1700
• B. 1650
• C. 3300
• D. 3500

Hint:

When dealing with AP problems, try to break the problem into parts by using the nth term formula and the sum of the first n terms formula.

Answer 1

The Correct Option is B

Let the first term of the AP be \( a \) and the common difference be \( d \). The formula for the nth term of an AP is given by: \[ T_n = a + (n-1) \cdot d \] The sum of the nth term and the mth term is: \[ T_{12} + T_{22} = 100 \] Substitute for \( T_{12} = a + 11d \) and \( T_{22} = a + 21d \): \[ (a + 11d) + (a + 21d) = 100 \] Simplify to: \[ 2a + 32d = 100 \] Now solve for \( a \) in terms of \( d \): \[ a + 16d = 50 \] \[ a = 50 - 16d \] Now, to find the sum of the first 33 terms of the AP, we use the formula for the sum of the first n terms: \[ S_n = \frac{n}{2} \cdot [2a + (n-1) \cdot d] \] For \( n = 33 \): \[ S_{33} = \frac{33}{2} \cdot [2a + 32d] \] Substitute \( a = 50 - 16d \): \[ S_{33} = \frac{33}{2} \cdot [2(50 - 16d) + 32d] \] Simplifying this expression: \[ S_{33} = \frac{33}{2} \cdot [100 - 32d + 32d] = \frac{33}{2} \cdot 100 = 1650 \]