## 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**= ****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 1,½,¼,⅛,1/16,……………………….

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)

**Following Question Prepared by Best Bank Coaching in Noida**

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

** Answer: 1**24

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 ?

** Answer: ** 81

In the sequence, all numbers are prime except 81

** 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/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

** 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 (upto 5 decimal)

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

5 + 55 + 555 + 5555 +…………………

**Answer: **5081(10n-1) –59n

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)10-1–59n

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

**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

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

**Answer:** 5/16

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

**Answer:** (364+ 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+………….+ 203is equal to

**Answer:** 24200

23+ 43+ 63+………….+ 203

= 8(13+ 23+ 33+ ………….+ 103)

= 8*3025

= 24200