Question

In an arithmetic progression, the sum of the first 10 terms is 210 and the sum of the next 10 terms is 610. What is the common difference of the arithmetic progression?

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

For any AP, if blocks of equal number of terms (n) are summed consecutively, the block sums themselves form a new Arithmetic Progression whose common difference is n^2d.

• Sum of first 10 terms (S_first 10) = 210

• Sum of next 10 terms (S_next 10) = 610
The difference between these two consecutive block sums is: 610 - 210 = 400. According to the property: n^2d = 400 (10)^2 d = 400 100d = 400 d = 4. This property avoids setting up large simultaneous equations entirely!

Answer 1

The Correct Option is C

Concept: The sum of the first n terms (S_n) of an Arithmetic Progression (AP) with first term a and common difference d is given by: \[ S_n = \frac{n}{2} \[2a + (n-1)d\] ]

Step 1: Construct the equation for the sum of the first 10 terms (S_10). We are given that S_10 = 210. Substituting n = 10 into the formula: \[ S_{10} = \frac{10}{2} \[2a + (10-1)d\] = 210 ] \[ 5 \[2a + 9d\] = 210 ] Divide both sides by 5: \[ 2a + 9d = 42 \quad \text{--- (Equation 1)} \]

Step 2: Construct the equation for the sum of the first 20 terms (S_20). The question states that the sum of the *next* 10 terms is 610. This means the total sum of the first 20 terms combined is: \[ S_{20} = S_{10} + \text{Sum of next 10 terms} = 210 + 610 = 820 \] Substituting n = 20 into the AP sum formula: \[ S_{20} = \frac{20}{2} \[2a + (20-1)d\] = 820 ] \[ 10 \[2a + 19d\] = 820 ] Divide both sides by 10: \[ 2a + 19d = 820 \div 10 \] \[ 2a + 19d = 822 \quad \text{--- (Equation 2)} \]

Step 3: Solve the system of linear equations to isolate the common difference d. Subtract Equation 1 from Equation 2 to eliminate the variable a: \[ (2a + 19d) - (2a + 9d) = 82 - 42 \] \[ 2a - 2a + 19d - 9d = 40 \] \[ 10d = 40 \] Divide through by 10: \[ d = \frac{40}{10} = 4 \] Thus, the common difference of the AP is 4, which aligns perfectly with Option (C).