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CAT 2025 Lesson : Progressions - Special Types I

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5. Special Types

5.1 Pattern for difference of squares, cubes, etc.



Pattern Example
1) Difference of consecutive terms in AP is constant
2) Difference of consecutive Squares will be in AP
3) Difference of difference of consecutive cubes will be in AP


Therefore, terms written in powers of consecutive integers can be reduced to Arithmetic Progression.

Example 19

Sum of the series 12−22+32−42+1^2 - 2^2 + 3^2 - 4^2 + ... +20012−20022+20032+ 2001^{2}-2002^{2}+ 2003^{2} is:

[IIFT 2008]

(1) 2007006    (2) 1005004    (3) 200506    (4) None of the above

Solution

The difference between the squares of consecutive integers is the sum of the two integers.

For instance, if xx and x+1x + 1 are the 22 consecutive integers,

(x+1)2−x2=2x+1(x + 1)^{2} - x^{2} = 2x+1 = (x+1)+x(x + 1) + x

Let S =12−22+32−42+1^2 - 2^2 + 3^2 - 4^2 + ... +20012−20022+20032+ 2001^{2}-2002^{2}+ 2003^{2}

= 12+(32−22)+(52−42)1^{2} + (3^{2}-2^{2}) + (5^{2}-4^{2}) + ... +(20032−20022)+ (2003^{2}-2002^{2})

= 1+5+9+...+40051 + 5 + 9 + ... + 4005

S is a series in arithmetic progression.

a=1,d=4a = 1,d = 4 and n=4005−14+1=1002n = \dfrac{4005 - 1}{4} + 1= 1002

S =n2×[2a+(n−1)d]\dfrac{n}{2}\times[2a + (n - 1)d] = 10022×(2+1001×4)\dfrac{1002}{2}\times(2 + 1001 \times 4)

=1002×2003=2007006=1002 \times 2003 = 2007006

Answer: 20070062007006

5.2 Arithmetico–Geometric Sequence


In this sequence, the numerators of the terms are in Arithmetic Progression, while the denominators of the terms are in Geometric Progression. Such questions need to be answered in the following manner.

Example 20

12+24+38+416+......=\dfrac{1}{2} + \dfrac{2}{4} + \dfrac{3}{8} + \dfrac{4}{16} +......=

Solution

As the differences of consecutive terms of an AP are constant,

Let x=12+222+323+424+....x = \dfrac{1}{2} + \dfrac{2}{2^2} + \dfrac{3}{2^3} + \dfrac{4}{2^4} +.... -----(1)

Now we multiply the terms by the common ratio of the GP in the denominator.

2x=1+22+322+423+524+.....2x =1 + \dfrac{2}{2} + \dfrac{3}{2^2} + \dfrac{4}{2^3} + \dfrac{5}{2^4}+..... -----(2)

Subtracting (1)(1) from (2)(2) ⇒ xx =1+12+122+123+124+.....1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \dfrac{1}{2^4}+.....

Here, we have a GP to infinity where a=1a = 1 and r=12r = \dfrac{1}{2}

x=a1−r=11−(1/2)x = \dfrac{a}{1 - r} = \dfrac{1}{1 - (1/2)} = 22

Answer: 22


In the following example, we have squares in the numerators. But, as discussed earlier, differences of consecutive squares will be in AP. We will, therefore, keep finding the differences till we get a constant term in the numerators, so that the sequence is in GP.

Example 21

12121+22122+32123+42124+....=?\dfrac{1^2}{12^1} + \dfrac{2^2}{12^2} + \dfrac{3^2}{12^3} + \dfrac{4^2}{12^4}+....=?

Solution

Let x=12121+22122+32123+42124+....x =\dfrac{1^2}{12^1} + \dfrac{2^2}{12^2} + \dfrac{3^2}{12^3} + \dfrac{4^2}{12^4}+.... -----(1)

Now we multiply the terms by the common ratio of the GP in the denominator.

12x=1+22121+32122+42123+52124+....12x =1 + \dfrac{2^2}{12^1} + \dfrac{3^2}{12^2} + \dfrac{4^2}{12^3} + \dfrac{5^2}{12^4}+.... -----(2)

Subtracting (1)(1) from (2)(2)

⇒ 11x=1+22−12121+32−22122+42−32123+52−42124+....11x =1 + \dfrac{2^2-1^2}{12^1} + \dfrac{3^2-2^2}{12^2} + \dfrac{4^2-3^2}{12^3} + \dfrac{5^2-4^2}{12^4}+....

⇒ 11x=1+3121+5122+7123+9124+......11x =1 + \dfrac{3}{12^1} + \dfrac{5}{12^2} + \dfrac{7}{12^3} + \dfrac{9}{12^4}+...... -----(3)

Once again, we multiply the terms by 1212.

⇒ 132x=12+3+5121+7122+9123+11124+......132x = 12 + 3 + \dfrac{5}{12^1} + \dfrac{7}{12^2} + \dfrac{9}{12^3} + \dfrac{11}{12^4}+...... -----(4)

⇒ (4)–(3)(4) – (3) ⇒ 121x=14+2121+2122+2123+2124121x = 14 + \dfrac{2}{12^1} + \dfrac{2}{12^2} + \dfrac{2}{12^3} + \dfrac{2}{12^4}     (a=212,r=112)\left( a = \dfrac{2}{12},r = \dfrac{1}{12} \right)

⇒ 121x=14+a1−r=14+2/121−(1/12)121x = 14 + \dfrac{a}{1 - r} = 14 + \dfrac{2/12}{1 - (1/12)} = 14+21114 + \dfrac{2}{11} = 15611\dfrac{156}{11}

⇒ 1561331\dfrac{156}{1331}

Answer: 1561331\dfrac{156}{1331}

5.3 Terms getting cancelled


Example 22

Wherexn=2n−12n+1x_n = \dfrac{2n - 1}{2n + 1} , what is x10×x11×x12×....x50x_{10} \times x_{11} \times x_{12}\times ....x_{50} ?

Solution

x10×x11×x12×....x50=1921×2123×2325×........99101×x_{10} \times x_{11} \times x_{12}\times ....x_{50} = \dfrac{19}{21}\times \dfrac{21}{23}\times \dfrac{23}{25}\times........\dfrac{99}{101}\times =19101\dfrac{19}{101}

(All terms, other than first terms's numerator and last term's denominator, will get cancelled.)

Answer:19101\dfrac{19}{101}

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