General Form of an AP: $$a, a + d, a + 2d, a + 3d, \dots$$ where $a$ is the first term and $d$ is the common difference.
Common Difference ($d$): $$d = a_{k+1} - a_k$$ Note that $d$ can be positive, negative, or zero.
For the AP: $3, 1, -1, -3, \dots$, write the first term $a$ and common difference $d$.
Solution:
First term $a = 3$
Common difference $d = a_2 - a_1$:
$$d = 1 - 3 = -2$$
Thus, $a = 3$ and $d = -2$.
$n^{\text{th}}$ Term from the End: $$a_n = l - (n - 1)d$$
Find the $10^{\text{th}}$ term of the AP: $2, 7, 12, \dots$
Solution:
Here $a = 2$, $d = 7 - 2 = 5$, and $n = 10$.
Using the formula:
$$a_n = a + (n - 1)d$$
$$a_{10} = 2 + (10 - 1)5$$
$$a_{10} = 2 + (9 \times 5)$$
$$a_{10} = 2 + 45$$
$$a_{10} = 47$$
The $10^{\text{th}}$ term is $47$.
Which term of the AP: $21, 18, 15, \dots$ is $-81$?
Solution:
Here $a = 21$, $d = 18 - 21 = -3$, and $a_n = -81$.
Substitute values into $a_n = a + (n - 1)d$:
$$-81 = 21 + (n - 1)(-3)$$
$$-81 - 21 = (n - 1)(-3)$$
$$-102 = (n - 1)(-3)$$
$$(n - 1) = \frac{-102}{-3}$$
$$n - 1 = 34$$
$$n = 35$$
Therefore, the $35^{\text{th}}$ term of the AP is $-81$.
Find the sum of the first $22$ terms of the AP: $8, 3, -2, \dots$
Solution:
Here $a = 8$, $d = 3 - 8 = -5$, and $n = 22$.
Using the sum formula:
$$S_n = \frac{n}{2} [2a + (n - 1)d]$$
$$S_{22} = \frac{22}{2} [2(8) + (22 - 1)(-5)]$$
$$S_{22} = 11 [16 + (21 \times -5)]$$
$$S_{22} = 11 [16 - 105]$$
$$S_{22} = 11 \times (-89)$$
$$S_{22} = -979$$
The sum of the first $22$ terms is $-979$.
How many terms of the AP: $24, 21, 18, \dots$ must be taken so that their sum is $78$?
Solution:
Here $a = 24$, $d = 21 - 24 = -3$, and $S_n = 78$.
Substitute into the formula:
$$S_n = \frac{n}{2} [2a + (n - 1)d]$$
$$78 = \frac{n}{2} [2(24) + (n - 1)(-3)]$$
$$156 = n [48 - 3n + 3]$$
$$156 = n [51 - 3n]$$
$$156 = 51n - 3n^2$$
Divide the equation by 3:
$$n^2 - 17n + 52 = 0$$
Factorising by splitting the middle term:
$$(n - 4)(n - 13) = 0$$
$$n = 4 \quad \text{or} \quad n = 13$$
Both values are acceptable because the sum of terms from $5^{\text{th}}$ to $13^{\text{th}}$ is zero. Therefore, $n$ can be $4$ or $13$.