General (nth) Term of AP
A pattern in which each term except the first is obtained by adding a fixed number (positive or negative) to the previous term is called an Arithmetic Progression (A.P.).
The first term is usually denoted by a and the common difference is denoted by d. Thus, standard form of an Arithmetic Progression is:
a, a + d, a + 2d, a + 3d, ...
General Term
Since first term is a, second term is obtained by adding d to a i.e., a + d, the third term will be obtained by adding d to a + d. So, third term will be (a + d) + d = a + 2d and so on.
First term, t1 = a
Second term, t2 = a + d
Third term, t3 = a + 2d
Fourth term, t4 = a + 3d
Each term is a + (term number - 1)d. So,
tn = a + (n - 1)d
Example 1: Find the 15th and nth terms of the AP: 16, 11, 6, 1, -4, -9, ...
Here, a = 16 and d = 11 - 16 = -5
t15 = a + (15 - 1)d = a + 14d
= 16 + 14(-5) = 16 0 70
= -54
Example 2: The first term of an AP is -3 and 12th term is 41. Determine the common difference.
Let first term of AP be a and common difference be d.
Therefore, t12 = a + (12 - 1)d = 41
-3 + 11d = 41
11d = 44
d = 4
Example 3: The common difference of an AP is 5 and 10th term is 43. Find its first term.
t10 = a + (10 - 1)d
43 = a + 9 × 5
43 = a + 45
a = -2
Example 4: The first term of an AP is -2 and 11th term is 18. Find its 15th term.
To find 15th term, you need to find d.
Now t11 = a + (11 - 1)d
18 = -2 + 10d
10d = 20
d = 2
Now t15 = a + 14d
= -2 + 14 × 2
= 26
Example 5: If p times the pth term of an AP is equal to q times the qth term, prove that its (p + q)th term is zero, provided p ≠ q.
tp = a + (p - 1)d
tq = a + (q - 1)d
Since p tp = q tq,
p[a + (p - 1)d] = q[a + (q - 1)d]
pa + p(p - 1)d - qa - q(q - 1)d = 0
(p - q)a + (p2 - q2)d - pd + qd = 0
(p - q)a + (p2 - q2)d - (p - q)d = 0
(p - q)a + (p - q)(p + q)d - (p - q)d = 0
(p - q)[a + (p + q)d - d] = 0
a + (p + q)d - d = 0 [as p - q ≠ 0]
a + (p + q - 1)d = 0
Since, LHS (p + q)th term, therefore, tp+q = 0