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## Sequence and Series Questions and Answers

Jul 31 • Bank, Maths Notes • 1755 Views • No Comments on Sequence and Series Questions and Answers

# Sequence and Series of Quantitative Aptitude

Arithmetic Progression: If the difference between any two consecutive terms of a sequence is same  then the sequence  is said to be arithmetic progression. It is denoted by A.P.

Example :  1,5,9,13,17,……………

-2,-5,-8,-11,-14,…………..

nth term or last term of an A.P. is given by the formula

an= a + (n-1) d

Where   a = first term of an A.P.,

d = common difference

n = number of terms

Arithmetic mean between any two given quantities a and b = (a+b)/2

Sum of first n terms of an A.P.

sn= n2 2a + (n-1)d or

sn = n2 2a +l

Where    l = the last term of an A.P.

Geometric progression: If the ratio of any two consecutive terms of a sequence is same then the sequence is said to be in geometric progression. It is denoted by G.P

Example 1,½,¼,⅛,1/16,……………………….

3, -9/4, 27/16, -81/16,…………….

nth term of a G.P. is given by the formula

.an= arn-1

Where      a = first term

R = common ratio

Sum of n terms of G.P.
sn= a (rn-1)/r-1 ,                    if r 1  and

sn= an ,                                  if r = 1

Sum of infinite terms  of a G.P. in case of  -1 < r <1

sn= a/(1-r)

Geometric mean between any two given quantities a and b = ab

Harmonic progression: a sequence a,b,c,d,…………is said to be in harmonic progression if 1/a,1/b,1/c,1/d,………………  is in A.P.

Example: ½,¼,⅙,……………..

Harmonic mean between any two given quantities a and b = 2ab/(a+b)

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Sum of special type of series:

• The sum of first n natural numbers i.e. 1+2+3+………+n = n (n+1)/2
• The sum of squares of first n natural numbers i.e.12+22+….+n2 = n(n+1)(2n+1)/6
• The sum of cubes of first  n natural numbers i.e. 13+23+…….+n3=n2(n+1)2/4
• The sum of first n odd natural numbers  =  n2
• The sum of first n even natural numbers = n2+ n
• The sum of squares of even natural numbers upto n2= n(n+1)(n+2)/6
• The sum of squares of odd natural numbers upto n2= n(n+1)(n+2)/6

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## Sequence and Series of Quantitative Aptitude WORKED -OUT PROBLEMS

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1. In the sequence of numbers 0, 7, 26, 63, ….., 215, 342 the missing term is

0 = 13-1

7 = 23-1

26 = 33-1

63 = 43-1

So missing number is = 53-1 = 124

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2) Which number in the sequence 41, 43, 47, 53, 61, 71, 73, 81, is wrongly written ?

In the sequence, all numbers are prime except 81

3) the sum (101 + 102 + 103 + …………..+ 200) is equal to

This is an A.P. with first term (a) = 101 , last term (l) = 200 and common difference = 1

an= a + (n-1)d

200 = 101 + n-1

n= 200-101+1 = 100

Sum = n/2a+l

= 100/2 101+200

= 50*301

= 15050

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4) If the 4th term of an arithmetic progression is 14 and the 12th term is 70, then the first

term is

4th term = a4= a + (4-1)d

= a + 3d = 14                                                                     (1)

12th term = a12= a + (12-1)d

= a + 11d = 70                                                                 (2)

Subtract (1) from (2) , we get

8d = 56

d = 7

Put the value of d in (1)

a + 21 = 14

a = -7

5) Find the value of  1 – 1/20 + 1/202– 1/203+……………………. Correct to 5  places of

Decimal is

This is an infinite  geometric progression with  a = 1 and  common ratio = -1/20

Sum = a/(1-r)

= 1/(1+1/20)

= 1/(21/20)

= 20/21

= 0.9523809

= 0.95238 (upto 5 decimal)

6) Find the sum of the first n terms of the series

5 + 55 + 555 + 5555 +…………………

5 + 55 + 555 + …………. upto n terms

5(1 + 11 + 111 + …………upto n terms )

59 (9 + 99 + 999 + ……….upto n terms)

59(10-1 + 100-1 + 1000-1 +…………..upto n terms)

59 (10 + 100 + 1000 +…………..upto n terms – n)

5910(10n-1)10-159n

5081(10n-1) –59n

7) When simplified, the sum  1/2  + 1/6  + 1/12 + 1/20 + 1/30 + ………. + 1/n(n+1) is equal to

1/2  + 1/6  + 1/12 + 1/20 + 1/30 + ………. + 1/n(n+1)

= 1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) +…………..+ 1/n(n+1)

= 1/1 – 1/2 + 1/2- 1/3  + 1/3 – 1/4 + 1/4 – 1/5 +………….+1/n – 1/ (n+1)

= 1-1/(n+1)

= (n+1-1)/(n+1)

= n/(n+1)

8) Find the sum of the following series
11*4+ 14*7+17*10+110*13+ 113*16

11*4+ 14*7+17*10+110*13+ 113*16

= 1331*4+ 34*7+37*10+310*13+ 313*16

=13(1 – 1/4  + 1/4  -1/7  + 1/7 – 1/10 + 1/10 – 1/13 + 1/13 – 1/16 )

= 13(1 – 1/16)

= 13(15/16)

= 5/16

9) 1 + (3 +1 ) (32+ 1)  (34+ 1)  (38+ 1)  (316+ 1)  (332+1) is equal to

1 + (3 +1 ) (32+ 1)  (34+ 1)  (38+ 1)  (316+ 1)  (332+1)

=1 + (3 – 1) (3 +1 ) (32+ 1)  (34+ 1)  (38+ 1)  (316+ 1)  (332+1)3-1

= 1 +  (3 – 1)(3 +1 ) (32+ 1)  (34+ 1)  (38+ 1)  (316+ 1)  (332+1)/2

= 1+(32– 1)  (32+ 1)  (34+ 1)  (38+ 1)  (316+ 1)  (332+1)/2

= 1+(34– 1) (34+ 1)  (38+ 1)  (316+ 1)  (332+1)/2

= 1 + (38– 1)  (38+ 1)  (316+ 1)  (332+1)/2

= 1 +(316– 1)  (316+ 1)  (332+1)/2

= 1 + (332-1) (332+1)/2

=1 + (364-1)/2

=( 2 + 364– 1 )/2

= (364+ 1)/2

10) Given that 13+ 23+ 33+ ………….+ 103= 3025, the value of 23+ 43+ 63+………….+ 203is equal to

23+ 43+ 63+………….+ 203

= 8(13+ 23+ 33+ ………….+ 103)

= 8*3025

= 24200