Progressions (or Sequences and Series) are numbers arranged in a particular order such that they form a predictable order. By predictable order, we mean that given some numbers, we can find next numbers in the series.
A sequence of numbers is called an arithmetic progression if the difference between any two consecutive terms is always same. In simple terms, it means that next number in the series is calculated by adding a fixed number to the previous number in the series. This fixed number is called the common difference.
For example, 2,4,6,8,10 is an AP because difference between any two consecutive terms in the series (common difference) is same (4 – 2 = 6 – 4 = 8 – 6 = 10 – 8 = 2).
If ‘a’ is the first term and ‘d’ is the common difference,
- nth term of an AP = a + (n-1) d
- Arithmetic Mean = Sum of all terms in the AP / Number of terms in the AP
- Sum of ‘n’ terms of an AP = 0.5 n (first term + last term) = 0.5 n [ 2a + (n-1) d ]
A sequence of numbers is called a geometric progression if the ratio of any two consecutive terms is always same. In simple terms, it means that next number in the series is calculated by multiplying a fixed number to the previous number in the series. This fixed number is called the common ratio.
For example, 2,4,8,16 is a GP because ratio of any two consecutive terms in the series (common difference) is same (4 / 2 = 8 / 4 = 16 / 8 = 2).
If ‘a’ is the first term and ‘r’ is the common ratio,
- nth term of a GP = a rn-1
- Geometric Mean = nth root of product of n terms in the GP
- Sum of ‘n’ terms of a GP (r < 1) = [a (1 – rn)] / [1 – r]
- Sum of ‘n’ terms of a GP (r > 1) = [a (rn – 1)] / [r – 1]
- Sum of infinite terms of a GP (r < 1) = (a) / (1 – r)
A sequence of numbers is called a harmonic progression if the reciprocal of the terms are in AP. In simple terms, a,b,c,d,e,f are in HP if 1/a, 1/b, 1/c, 1/d, 1/e, 1/f are in AP.
For two terms ‘a’ and ‘b’,
- Harmonic Mean = (2 a b) / (a + b)
For two numbers, if A, G and H are respectively the arithmetic, geometric and harmonic means, then
- A ≥ G ≥ H
- A H = G2, i.e., A, G, H are in GP
Question 1 : Find the nth term for the AP : 11, 17, 23, 29, …
Solution : Here, a = 11, d = 17 – 11 = 23 – 17 = 29 – 23 = 6
We know that nth term of an AP is a + (n – 1) d
=> nth term for the given AP = 11 + (n – 1) 6
=> nth term for the given AP = 5 + 6 n
We can verify the answer by putting values of ‘n’.
=> n = 1 -> First term = 5 + 6 = 11
=> n = 2 -> Second term = 5 + 12 = 17
=> n = 3 -> Third term = 5 + 18 = 23
and so on …
Question 2 : Find the sum of the AP in the above question till first 10 terms.
Solution : From the above question,
=> nth term for the given AP = 5 + 6 n
=> First term = 5 + 6 = 11
=> Tenth term = 5 + 60 = 65
=> Sum of 10 terms of the AP = 0.5 n (first term + last term) = 0.5 x 10 (11 + 65)
=> Sum of 10 terms of the AP = 5 x 76 = 380
Question 3 : For the elements 4 and 6, verify that A ≥ G ≥ H.
Solution : A = Arithmetic Mean = (4 + 6) / 2 = 5
G = Geometric Mean = = 4.8989
H = Harmonic Mean = (2 x 4 x 6) / (4 + 6) = 48 / 10 = 4.8
Therefore, A ≥ G ≥ H
Question 4 : Find the sum of the series 32, 16, 8, 4, … upto infinity.
Solution : First term, a = 32
Common ratio, r = 16 / 32 = 8 / 16 = 4 / 8 = 1 / 2 = 0.5
We know that for an infinite GP, Sum of terms = a / (1 – r)
=> Sum of terms of the GP = 32 / (1 – 0.5) = 32 / 0.5 = 64
Question 5 : The sum of three numbers in a GP is 26 and their product is 216. ind the numbers.
Solution : Let the numbers be a/r, a, ar.
=> (a / r) + a + a r = 26
=> a (1 + r + r2) / r = 26
Also, it is given that product = 216
=> (a / r) x (a) x (a r) = 216
=> a3 = 216
=> a = 6
=> 6 (1 + r + r2) / r = 26
=> (1 + r + r2) / r = 26 / 6 = 13 / 3
=> 3 + 3 r + 3 r2 = 13 r
=> 3 r2 – 10 r + 3 = 0
=> (r – 3) (r – (1 / 3) ) = 0
=> r = 3 or r = 1 / 3
Thus, the required numbers are 2, 6 and 18.
This article has been contributed by Nishant Arora
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