Arithmetic Progressions
This chapter covers arithmetic progressions (AP), finding the nth term and the sum of the first n terms of an AP, and solving real-life problems using APs.
An arithmetic progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a fixed number (called the common difference, d) to the preceding term. For example: 2, 5, 8, 11, ... has common difference 3.
The nth term of an AP is: an = a + (n-1)d, where a is the first term and d is the common difference. The sum of the first n terms is: Sn = n/2 [2a + (n-1)d], or equivalently Sn = n/2 (a + an), where an is the last term.
Q. Find the 10th term of the AP 2, 7, 12, 17, ...
Here a = 2, d = 7 - 2 = 5. Using an = a + (n-1)d: a10 = 2 + (10-1)(5) = 2 + 45 = 47.
In an AP, if a, b, c are three consecutive terms, then b is the arithmetic mean of a and c, i.e., b = (a + c) / 2.
Q. Find the sum of the first 15 terms of the AP 3, 7, 11, 15, ...
Here a = 3, d = 4, n = 15. Sn = n/2 [2a + (n-1)d] = 15/2 [2(3) + 14(4)] = 15/2 [6 + 56] = 15/2 (62) = 465.
The common difference of the AP -5, -1, 3, 7, ... is:
Correct answer: (C) 4
In an AP with first term a and common difference d, if d is negative, the AP is said to be _______.
Decreasing
The common difference of an AP can be a negative number, zero, or a positive number.
True — the common difference can be any real number: positive (increasing AP), negative (decreasing AP), or zero (a constant sequence, which is technically also an AP).
How many terms of the AP 9, 17, 25, ... must be taken to give a sum of 636?
Here a = 9, d = 8. Using Sn = n/2[2a + (n-1)d] = 636: n/2[18 + 8(n-1)] = 636, so n[18 + 8n - 8] = 1272, giving n(8n + 10) = 1272, or 8n^2 + 10n - 1272 = 0, which simplifies to 4n^2 + 5n - 636 = 0. Solving this quadratic: n = 12 (taking the positive root). So 12 terms are needed.