Arithmetic Progressions | Exercise 5.2

Question 12

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution

Understand the Question

The $n$ -th term of an arithmetic progression (AP) is given by $a_n = a + (n - 1)d$ .
When two APs share the same common difference $d$ , the difference between their $n$ -th terms is: $(a + (n - 1)d) - (b + (n - 1)d) = a - b$
Because the $(n - 1)d$ terms always cancel out, the difference between any corresponding terms of the two APs is constant and strictly equal to the difference between their first terms ( $a - b$ ).

Step 1 · Define the Terms of Both APs

Let the first term of the first AP be $a_1$ and that of the second AP be $b_1$ . Let the common difference for both APs be $d$ . Diagram 1

The $n$ -th term of the first AP is: $A_n = a_1 + (n - 1)d$

The $n$ -th term of the second AP is: $B_n = b_1 + (n - 1)d$

Step 2 · Find the Difference Between the 100th Terms

For the 100th terms:

\begin{aligned} A_{100} &= a_1 + (100 - 1)d \\[0.6em] &= a_1 + 99d \end{aligned}

\begin{aligned} B_{100} &= b_1 + (100 - 1)d \\[0.6em] &= b_1 + 99d \end{aligned}

Subtracting the two terms:

\begin{aligned} A_{100} - B_{100} &= (a_1 + 99d) - (b_1 + 99d) \\ &= a_1 + 99d - b_1 - 99d \\ &= a_1 - b_1 \end{aligned}

Given that $A_{100} - B_{100} = 100$ : $a_1 - b_1 = 100 \quad \dots (1)$

Step 3 · Calculate the Difference Between the 1000th Terms

For the 1000th terms:

\begin{aligned} A_{1000} &= a_1 + (1000 - 1)d \\[0.6em] &= a_1 + 999d \end{aligned}

\begin{aligned} B_{1000} &= b_1 + (1000 - 1)d \\[0.6em] &= b_1 + 999d \end{aligned}

Subtracting the two terms:

\begin{aligned} A_{1000} - B_{1000} &= (a_1 + 999d) - (b_1 + 999d) \\ &= a_1 + 999d - b_1 - 999d \\ &= a_1 - b_1 \end{aligned}

From equation $(1)$ , $a_1 - b_1 = 100$ : $A_{1000} - B_{1000} = 100$

Answer

$100$

Common Mistakes

Assuming Proportional Growth: Incorrectly multiplying the difference by $10$ (e.g., guessing $1000$ because $1000 = 10 \times 100$ ). The difference depends only on the first terms, not $n$ .
Sign Errors in Subtraction: Forgetting to distribute the negative sign across parentheses, writing $-(b_1 + 99d)$ as $-b_1 + 99d$ instead of $-b_1 - 99d$ .

More questions in Exercise 5.2