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An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference.
If the first term of an AP is \( a \) and the common difference is \( d \), then the sequence is:
\[a, \quad a + d, \quad a + 2d, \quad a + 3d, \quad \ldots\]Here, each term increases (or decreases) by \( d \) from the previous term.
The common difference is given by:
\[d = t_{n} - t_{n-1}\]where \( t_n \) is the \( n^{th} \) term.
The \( n^{th} \) term of an AP, denoted by \( t_n \), is given by:
\[t_n = a + (n - 1)d\]This formula helps find any term in the sequence without listing all previous terms.
The sum of the first \( n \) terms of an AP is denoted by \( S_n \). There are two commonly used formulas:
where \( l = t_n \) is the last term.
Both formulas are equivalent and can be used depending on the known information.
Arithmetic progressions appear in many real-life situations such as:
Find the 10th term of an AP whose first term is 3 and common difference is 5.
Solution:
Given: \( a = 3 \), \( d = 5 \), \( n = 10 \)
Using the nth term formula:
\[ t_{10} = a + (10 - 1)d = 3 + 9 \times 5 = 3 + 45 = 48 \]Answer: The 10th term is 48.
Find the sum of the first 20 terms of an AP with first term 7 and common difference 3.
Solution:
Given: \( a = 7 \), \( d = 3 \), \( n = 20 \)
First find the 20th term:
\[ t_{20} = a + (20 - 1)d = 7 + 19 \times 3 = 7 + 57 = 64 \]Sum of first 20 terms:
\[ S_{20} = \frac{20}{2} (7 + 64) = 10 \times 71 = 710 \]Answer: The sum is 710.
Find the number of terms in an AP if the first term is 5, common difference is 2, and the last term is 65.
Solution:
Given: \( a = 5 \), \( d = 2 \), \( l = 65 \)
Using the nth term formula:
\[ l = a + (n - 1)d \implies 65 = 5 + (n - 1) \times 2 \] \[ 65 - 5 = 2(n - 1) \implies 60 = 2(n - 1) \implies n - 1 = 30 \implies n = 31 \]Answer: The number of terms is 31.
The sum of the first \( n \) terms of an AP is given by \( S_n = 3n^2 + 2n \). Find the first term and the common difference.
Solution:
We know:
\[ S_n = \sum_{k=1}^n t_k = 3n^2 + 2n \]The \( n^{th} \) term is:
\[ t_n = S_n - S_{n-1} = [3n^2 + 2n] - [3(n-1)^2 + 2(n-1)] \] \[ = 3n^2 + 2n - [3(n^2 - 2n + 1) + 2n - 2] \] \[ = 3n^2 + 2n - [3n^2 - 6n + 3 + 2n - 2] = 3n^2 + 2n - 3n^2 + 6n - 3 - 2n + 2 \] \[ = (2n + 6n - 2n) + (-3 + 2) = 6n - 1 \]So, \( t_n = 6n - 1 \).
Find \( t_1 \):
\[ t_1 = 6 \times 1 - 1 = 5 \]Find common difference \( d = t_2 - t_1 \):
\[ t_2 = 6 \times 2 - 1 = 11, \quad d = 11 - 5 = 6 \]Answer: First term \( a = 5 \), common difference \( d = 6 \).
Find the sum of all two-digit numbers divisible by 7.
Solution:
Two-digit numbers divisible by 7 start from 14 and end at 98.
So, the AP is:
\[ 14, 21, 28, \ldots, 98 \]Here, \( a = 14 \), \( d = 7 \), last term \( l = 98 \).
Find number of terms \( n \):
\[ l = a + (n - 1)d \implies 98 = 14 + (n - 1) \times 7 \] \[ 98 - 14 = 7(n - 1) \implies 84 = 7(n - 1) \implies n - 1 = 12 \implies n = 13 \]Sum of these terms:
\[ S_n = \frac{n}{2} (a + l) = \frac{13}{2} (14 + 98) = \frac{13}{2} \times 112 = 13 \times 56 = 728 \]Answer: The sum is 728.
| Concept | Formula | Remarks |
|---|---|---|
| General term (nth term) | \( t_n = a + (n - 1)d \) | Find any term in the AP |
| Sum of first n terms (using last term) | \( S_n = \frac{n}{2} (a + l) \) | Use when last term \( l \) is known |
| Sum of first n terms (using common difference) | \( S_n = \frac{n}{2} \left[ 2a + (n - 1)d \right] \) | Use when \( d \) is known but last term is unknown |
| Common difference | \( d = t_n - t_{n-1} \) | Difference between consecutive terms |
| Relation between three consecutive terms | \( t_k = \frac{t_{k-1} + t_{k+1}}{2} \) | Middle term is average of neighbors |
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