CBSE Class 10 Mathematics Arithmetic Progressions Notes Set 13

What Is an Arithmetic Progression

A list of numbers in which each term after the first is obtained by adding a fixed number d to the previous term is called an arithmetic progression (AP). The fixed number d is the common difference; it can be positive, negative or zero. The first term is a.
General form: a, a + d, a + 2d, a + 3d, ...
To test whether a list is an AP, check that a₂ − a₁ = a₃ − a₂ = a₄ − a₃ = ... If the differences are not all equal, it is not an AP.

The nth Term

aₙ = a + (n − 1)d
The nth term from the end of a finite AP with last term l is l − (n − 1)d.
Three numbers a, b, c are in AP exactly when 2b = a + c, that is, b is the arithmetic mean of a and c.
If aₙ is a linear expression in n, the sequence is an AP and the coefficient of n is the common difference.

Sum of the First n Terms

Sₙ = \( \frac{n}{2}[2a + (n − 1)d] \) = \( \frac{n}{2}(a + l) \), where l is the last term.
The nth term can be found from sums: aₙ = Sₙ − Sₙ₋₁ (for n ≥ 2) and a₁ = S₁.
Sum of the first n natural numbers = n(n + 1)/2, sum of the first n odd numbers = n², sum of the first n even numbers = n(n + 1).

Useful Tricks for Choosing Terms

1. For three terms in AP take a − d, a, a + d (their sum is 3a).
2. For four terms take a − 3d, a − d, a + d, a + 3d (common difference 2d).
3. For five terms take a − 2d, a − d, a, a + d, a + 2d.
4. In an AP the sum of terms equidistant from the beginning and the end is the same.
5. If Sₚ = Sq (p ≠ q), then S₍ₚ₊q₎ = 0. If Sₚ = q and Sq = p, then S₍ₚ₊q₎ = −(p + q).
6. For an AP, aₘ = n and aₙ = m gives a₍ₘ₊ₙ₎ = 0.

Worked Examples

Example 1. Which term of the AP 3, 8, 13, 18, ... is 78?
Solution: a = 3, d = 5. 3 + (n − 1) × 5 = 78 gives n − 1 = 15, so n = 16. The 16th term is 78.

Example 2. Find the sum of the first 22 terms of the AP 8, 3, −2, ...
Solution: a = 8, d = −5, n = 22. S = 22/2 × [16 + 21 × (−5)] = 11 × (−89) = −979.

Example 3. The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.
Solution: a + 7d = 0, so a = −7d. a₁₈ = a + 17d = 10d and a₃₈ = a + 37d = 30d = 3 × a₁₈.

Example 4. The sum of the first n terms of an AP is 3n² + 5n. Find the AP and its 12th term.
Solution: S₁ = 8 = a₁. S₂ = 22, so a₂ = 14, d = 6. The AP is 8, 14, 20, ... and a₁₂ = 8 + 11 × 6 = 74.

Example 5. Find the sum of all two-digit numbers divisible by 7.
Solution: The numbers are 14, 21, ..., 98, so n = 13. Sum = 13/2 × (14 + 98) = 728.

Example 6. A sum of Rs. 700 is to be used to give seven cash prizes, each prize Rs. 20 less than the preceding one. Find the value of each prize.
Solution: 7/2 × (2a − 120) = 700, so 2a − 120 = 200 and a = 160. The prizes are Rs. 160, 140, 120, 100, 80, 60 and 40.