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

Step 1: Recall the AP sum formula.
The sum of the first $n$ terms of an arithmetic progression is $S_n = \frac{n}{2}\big[2a + (n-1)d\big]$, where $a$ is the first term and $d$ is the common difference.

Step 2: Use the first ten terms.
With $S_{10} = 210$ and $n = 10$, \[ \frac{10}{2}[2a + 9d] = 210 \;\Rightarrow\; 2a + 9d = 42 \]

Step 3: Build the sum of twenty terms.
The next ten terms add $610$, so the first twenty together give $210 + 610 = 820$. So $S_{20} = 820$.

Step 4: Write the second equation.
With $n = 20$, \[ \frac{20}{2}[2a + 19d] = 820 \;\Rightarrow\; 2a + 19d = 82 \]

Step 5: Subtract the equations.
Take the first from the second to remove $a$. \[ (2a + 19d) - (2a + 9d) = 82 - 42 \;\Rightarrow\; 10d = 40 \]

Step 6: Solve for $d$.
Divide by $10$ to get $d = 4$. So the common difference is \[ \boxed{4} \]