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# Sequence and Series Questions for Bank Po

Sequence and Series Questions for Bank Po, IBPS, SBI  government exam

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

a= 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.

s= n/2 [2a + (n-1)d] or

s = n/2 [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 1,1/2, 1/4, 1/8,1/16,……………………….

3, -9/4, 27/16, -81/16,…………….

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

a= arn-1

Where      a = first term

R = common ratio

## Sequence and Series Questions for Bank Po

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

s= 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

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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+….+n= n (n+1) (2n+1)/6
• The sum of cubes of first  n natural numbers i.e. 12+…….+n= n(n+1) 2/4
• The sum of first n odd natural numbers  = n2
• The sum of first n even natural numbers = n+ n
• The sum of squares of even natural numbers up to n= n (n+1)(n+2)/6
• The sum of squares of odd natural numbers up to n= n (n+1)(n+2)/6

WORKED -OUT PROBLEMS

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

0 = 1– 1

7 = 2– 1

26 = 3– 1

63 = 4– 1

So missing number is = 5– 1 = 124

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

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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/2 [a+l]

= 100/2 [101+200]

= 50*301

= 15050

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

4th term = a= 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 + 1/20– 1/20+……………………. 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 (up to 5 decimal)

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

5 + 55 + 555 + 5555 +…………………

Answer: 50/81 × (10– 1) –5/9n

5 + 55 + 555 + …………. up to n terms

5 (1 + 11 + 111 + …………up to n terms )

5/9 (9 + 99 + 999 + ……….up to n terms)

5/(10-1 + 100-1 + 1000 – 1 +…………..up to n terms)

5/9 (10 + 100 + 1000 +…………..up to n terms – n)

5/9 × 10 (10– 1) 10 – 1 – 59n

50/81 × (10– 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
1/1*4+ 1/4*7+1/7*10+1/10*13+ 1/13*16

1/1*4+ 1/4*7+1/7*10+1/10*13+ 1/13*16

= 1/3[3/1*4+ 3/4*7+3/7*10+3/10*13+ 3/13*16]

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

= 1/3(1 – 1/16)

= 1/3(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+………….+ 203 is equal to

23+ 43+ 63+………….+ 203

= 8 (1+ 2+ 3+ ………….+ 103)

= 8*3025

= 24200

1. 9  23  75  ?  1543  9267

2. 253                 b. 307           c. 356               d. 411              e. 457