CBSE Class 10 Mathematics Arithmetic Progressions Notes Set 12

Sequences and Arithmetic Progression

A sequence is a list of numbers in a definite order. An A.P. has a constant difference d between consecutive terms: aₙ = a + (n − 1)d.
Sum of n terms: Sₙ = (n/2)[2a + (n − 1)d] = (n/2)(a + l), where l is the last term. The nth term can be found from sums as aₙ = Sₙ − Sₙ₋₁.
Three numbers in A.P. can be taken as a − d, a, a + d; four numbers as a − 3d, a − d, a + d, a + 3d. The arithmetic mean of a and b is (a + b)/2.

Geometric Progression

A G.P. has a constant ratio r: aₙ = arⁿ⁻¹. Sum of n terms: Sₙ = a(rⁿ − 1)/(r − 1) for r ≠ 1.
If |r| < 1 the infinite G.P. has sum S∞ = a/(1 − r). The geometric mean of a and b is √(ab). Three numbers in G.P. can be taken as a/r, a, ar.

Harmonic Progression and the Three Means

Numbers are in H.P. when their reciprocals are in A.P., so the nth term of an H.P. is 1/[a + (n − 1)d] where a and d belong to the reciprocal A.P. The harmonic mean of a and b is 2ab/(a + b).
For positive numbers, A.M. ≥ G.M. ≥ H.M., with equality only when all the numbers are equal. For two numbers, G.M.² = A.M. × H.M.

Useful Sums

1 + 2 + ... + n = n(n + 1)/2. 1² + 2² + ... + n² = n(n + 1)(2n + 1)/6. 1³ + 2³ + ... + n³ = [n(n + 1)/2]².
Telescoping: a fraction of the form k/[n(n + k)] can be written as 1/n − 1/(n + k), so most terms cancel.

Solved Examples

Example 1. Find the 20th term of the A.P. 5, 8, 11, ...
Answer:
a = 5, d = 3, so t₂₀ = 5 + 19 × 3.
62

Example 2. How many terms of 3, 7, 11, ... add up to 465?
Answer:
(n/2)[6 + 4(n − 1)] = 465 gives n(2n + 1) = 465, i.e. 2n² + n − 465 = 0.
(2n + 31)(n − 15) = 0, so n = 15.

Example 3. The 3rd term of an A.P. is 8 and the 7th term is 20. Find the sum of the first 10 terms.
Answer:
a + 2d = 8 and a + 6d = 20 give d = 3, a = 2.
S₁₀ = 5(4 + 27) = 155

Example 4. Find the sum of all multiples of 3 between 10 and 100.
Answer:
The numbers 12, 15, ..., 99 form an A.P. with 30 terms.
Sum = 15 × (12 + 99) = 1665

Example 5. Find the sum of the first 8 terms of the G.P. 2, 6, 18, ...
Answer:
a = 2, r = 3: S₈ = 2(3⁸ − 1)/2 = 3⁸ − 1.
6560

Example 6. Find the sum of the infinite series 6 + 2 + 2/3 + ...
Answer:
r = 1/3, so S = 6/(1 − 1/3).
9

Example 7. Insert two geometric means between 3 and 81.
Answer:
81 = 3r³, so r³ = 27 and r = 3.
The means are 9 and 27.

Example 8. The A.M. of two numbers is 25 and their G.M. is 15. Find the numbers and their H.M.
Answer:
a + b = 50 and ab = 225, so the numbers satisfy t² − 50t + 225 = 0, i.e. (t − 5)(t − 45) = 0.
The numbers are 5 and 45, and H.M. = 2 × 225/50 = 9.

Example 9. Find 1/(1·2) + 1/(2·3) + ... + 1/(9·10).
Answer:
Each term is 1/n − 1/(n + 1), so the sum telescopes to 1 − 1/10.
9/10

Example 10. If 5th term of an H.P. is 1/10 and the 10th term is 1/20, find the 15th term.
Answer:
The reciprocals form an A.P.: 10 and 20 are the 5th and 10th terms, so d = 2 and the 15th term is 30.
1/30