How To Solve Arithmetic Progression Questions Quickly

Type 3: Sum of first ‘n’ terms of the series: S_{n} = \frac{n}{2}\left [ 2a + (n-1) d\right ] \text{ or } \frac{n}{2}(a + l)

Question 1. Sanjana has one paisa on 1^st of January with her. She got 2 paisa on the 2^nd January and 3 paisa on 3^rd of January. In similar fashion, what will be her saving after 365 days?

Options:

A. 66700 paisa

B. 66795 paisa

C. 65795 paisa

D. 66790 paisa

Solution:     According to the question, the series formed is 1,2,3,4,5,……

Therefore, a = 1 and d = 1

S_{n} = \frac{n}{2}\left [ 2a + (n-1) d\right ]

n = 365

S₃₆₅ =  \frac{365}{2}\times [2 \times 1 + (365 − 1) \times 1]

S₃₆₅ = 66795 paisa

Correct option: C

Question 2.Find the sum of the series 62 + 60 + 58 …….. + 30

Options:

A. 728

B.780

C.719

D. 782

Solution:    In the given series a = 62,  d = -2,  t_n = 30

t_n = a + (n – 1)d

30= 62 + (n – 1)-2

30 = 62 – 2n + 2

2n = 62-30+2

34 = 2n

n = 17

S_{n} = \frac{n}{2}(a + l)

S₁₇ = \frac{17}{2}\times (62 + 30)

S₁₇ =  \frac{17}{2}\times (92)

S₁₇ = 782

Correct option: D

Question 3.If the sum of 5 consecutive numbers is 100 then find out the first number.

Options:

A. 20

B. 25

C. 18

D. 16

Solution:     It is clear from the question that d = 1 because there are 5 consecutive numbers

n = 5

S_n= 100

Let the first number ‘a’ be x

S_{n} = \frac{n}{2}\left [ 2a + (n-1) d\right ]

100 = \frac{5}{2} [2a + (5 − 1) \times 1]

100 =  \frac{5}{2}[2a + (4) \times 1]

100\times \frac{2}{5} =  [2a + 4]

40 = 2a + 4

a = 18

So, the first number is 18 and the complete series is 18, 19, 20, 21, and 22.

Correct option: C