Sequence and Series Questions for Bank Po
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
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 = n/2 [2a + (n-1)d] or
sn = 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
an = arn-1
Where a = first term
R = common ratio
Sequence and Series Questions for Bank Po
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
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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+….+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 up to n2 = n (n+1)(n+2)/6
- The sum of squares of odd natural numbers up to n2 = n (n+1)(n+2)/6
WORKED -OUT PROBLEMS
- In the sequence of numbers 0, 7, 26, 63, ….., 215, 342 the missing term is
Answer: 124
0 = 13 – 1
7 = 23 – 1
26 = 33 – 1
63 = 43 – 1
So missing number is = 53 – 1 = 124
2) Which number in the sequence 41, 43, 47, 53, 61, 71, 73, 81, is wrongly written ?
Answer: 81
In the sequence, all numbers are prime except 81
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3) the sum (101 + 102 + 103 + …………..+ 200) is equal to
Answer : 15050
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
Answer: -7
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
Answer: 0.95238
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 × (10n – 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)
5/9 (10-1 + 100-1 + 1000 – 1 +…………..up to n terms)
5/9 (10 + 100 + 1000 +…………..up to n terms – n)
5/9 × 10 (10n – 1) 10 – 1 – 59n
50/81 × (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
Answer: n/(n+1)
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
Answer: 5/16
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
Answer: (332+ 1)/2
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
Answer: 24200
23+ 43+ 63+………….+ 203
= 8 (13 + 23 + 33 + ………….+ 103)
= 8*3025
= 24200
- 9 23 75 ? 1543 9267
2. 253 b. 307 c. 356 d. 411 e. 457
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