Tips on cracking Aptitude Questions on Progressions

1. 4 TIPS on cracking Aptitude Questions on Progressions

2. 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.

3. 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.

4. 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 Tn= 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 rth term of the series can be written as Tr= 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.

5. 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 3rd 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.

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