4 TIPS on cracking Aptitude Questions on Progressions

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Tip #1: Sum of ‘n’ terms of an AP= n x (Arithmetic Mean of first and last terms)

If a be the first term of an AP and l be the last term, i.e., the nth term, then the sum of the AP will be n(a + l)/2.

If the last term is not specified, replace l by a + (n – 1)d where d is the common difference. Then the sum of the AP will be n {2a + (n – 1)d}/2

Question: The interior angles of a polygon are in AP. The smallest angle is 120̊ and the common difference is 5̊. Find the number of sides of the polygon.

Solution:

Let there be n sides.

Then the largest side is 120 + (n – 1)5 degrees= 115̊ + 5n ̊

Sum of interior angles= [n(120 + 115 + 5n)/2] ̊ = [n(235 + 5n)/2] ̊

In any polygon, the sum of the interior angles is (n – 2) x 180 ̊

=>   [n(235 + 5n)/2] ̊ = (n – 2) x 180 ̊

=>   n² – 25n + 144=0

=>   (n – 9)(n – 16) = 0

=>    n= 9 or 16

If n=16, largest angle= 115 ̊ + 5̊ x16 = 195̊

But the interior angle of a polygon cannot be greater than 180 ̊.

Hence, n= 9.

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Tip #2: If the terms of an AP are operated upon by some constant, the resultant AP bears a certain relation to the previous one

If each term of an AP is multiplied by C, common difference becomes dC and the sum SC

If a constant C is added to each term of an AP the new sum is S+nC and d remains same

Question: 6 kids are born into a family and their age difference is 3 years. If the current age of the youngest child is 4 years, what will be the sum of their ages 5 years from now?

Solution:

According to the question, age of the eldest child = 4 + 5 * 3 = 19 yrs

Sum of their ages now= 6(19 + 4)/2= 69 years.

Sum of their ages 5 years from now= 69 + (6 x 5) = 99 years.

Question: The sum of 10 numbers in an AP is 810. What is their sum when all the numbers are divided by 3?

Solution:

New sum= 810/3= 270.

Note: Similarly, if each term of an AP is divided by a constant C, then the common difference becomes d/C and the sum of the new series is S/C. Again, if a constant C is subtracted from each term, then the common difference remains unchanged and the new sum is S – nC.

Tip #3: Wherever possible, use the Formulae for Sum

Sum of first n natural numbers, 1, 2, 3, …., n is: S= n(n+1)/2

Sum of squares of first n natural numbers 1², 2², 3², …., n² is: S= n(n+1)(2n+1)/6

Sum of cubes of first n natural numbers, 1³, 2³, 3³, …., n³ is: S= n²(n+1)²/4

If the nth term of a sequence is T_n= an³ + bn² + cn + d, then the sum of the series is: S= aΣn³ + bΣn² + cΣⁿ + nd

Question: Find the sum of n + 2(n – 1) + 3(n – 2) + …. + (n – 1)2 + n

Solution:

The r^th term of the series can be written as T_r = r(n – r + 1)

                                                                   = (n+1)r – r².

Thus, the sum of the series will be:

S= (n+1)Σr – Σr²

  = (n+1)Σn – Σn²

  = (n+1).n.(n+1)/2 – n(n+1)(2n+1)/6

  = n(n+1)(3n+3 – 2n – 1)/6

  = n(n+1)(n+2)/6.

Note: Most of the times, the sums can be calculated using some predefined formulae to save both time and effort for calculations. This is useful not only in questions on Progressions, but also helps in all types of question.

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Tip #4: When the quantities are in progression, choose the variable as the middle element to make your calculations easier

Question: The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?

Solution:

Let the age of the 3^rd child be a.

Then the successive ages of the children are (a-6), (a-3), a, (a+3), (a+6).

According to the question,

5a=50             (Since the positive and negative terms are exactly equal and thus cancel out.)

=>   a=10

Age of youngest child= a-6= 4 years

Question: A, B and C are neighbors. If A’s age is thrice that of B and B’s age is thrice that of C and the product of their ages is 216, what is C’s age?

Solution:

Let B’s age be b. Then,

=>   3b * b * (b/3) = 216

=>   b³ = 216   (Since the multipliers and divisors cancel each other)

=>   b= (216)^1/3 = 6

=>    C’s age= b/3= 2 years.

Note: In case the quantities mentioned form an arithmetic or geometric progression, choose the middle element as x so that the equation is reduced to being as a multiple or exponent of x. Thus, for an AP of 4 terms, let the terms be (a -3d), (a – d), (a + d), (a + 3d) and for an AP of 5 terms, the terms would be (a – 2d), (a – d), a, (a + d), (a + 2d). Similarly, for a GP of 4 terms, let the terms be a/r³, a/r, ar, ar³ and for a GP of 5 terms, take them as a/r², a/r, a, ar, ar²

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