Question

The 7th term of an Arithmetic Progression is 40 and the 13th term is 70. Find the 21st term of the AP.

• A. 100
• B. 110
• C. 120
• D. 130

Hint:

You can find $d$ instantly without writing equations! The difference between terms divided by the difference in their positions gives $d$: $$d = \frac{a_{13} - a_{7}}{13 - 7} = \frac{70 - 40}{6} = 5$$

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
An Arithmetic Progression (AP) is a sequence where each term is obtained by adding a fixed value, called the common difference ($d$), to the preceding term. Given any two terms, we can build a system of linear equations to solve for the first term ($a$) and the common difference ($d$).

Step 2: Key Formula or Approach:
The $n$-th term ($a_n$) of an AP is given by the formula: $$a_n = a + (n - 1)d$$ Where: $a = \text{first term}$ $d = \text{common difference}$

Step 3: Detailed Explanation:
From the problem parameters, we write the expressions for both given terms: 1. For the 7th term ($a_7 = 40$): \[ a + (7 - 1)d = 40 \implies a + 6d = 40 \quad \text{--- (Equation 1)} \] 2. For the 13th term ($a_{13} = 70$): \[ a + (13 - 1)d = 70 \implies a + 12d = 70 \quad \text{--- (Equation 2)} \] Subtract Equation 1 from Equation 2 to eliminate $a$: \[ (a + 12d) - (a + 6d) = 70 - 40 \] \[ 6d = 30 \implies d = \frac{30}{6} = 5 \] Substitute $d = 5$ back into Equation 1 to find $a$: \[ a + 6(5) = 40 \] \[ a + 30 = 40 \implies a = 10 \] Now, calculate the 21st term ($a_{21}$) using our values $a = 10$ and $d = 5$: \[ a_{21} = a + (21 - 1)d \] \[ a_{21} = 10 + 20(5) \] \[ a_{21} = 10 + 100 = 110 \]

Step 4: Final Answer:
The 21st term of the AP is 110.