Answer:
The sum of the first 100 terms is 60400
Step-by-step explanation:
* Lets revise the arithmetic sequence
- There is a constant difference between each two consecutive
numbers
- Ex:
# 2 , 5 , 8 , 11 , ……………………….
# 5 , 10 , 15 , 20 , …………………………
# 12 , 10 , 8 , 6 , ……………………………
* General term (nth term) of an Arithmetic sequence:
- U1 = a , U2 = a + d , U3 = a + 2d , U4 = a + 3d , U5 = a + 4d
- Un = a + (n – 1)d, where a is the first term , d is the difference
between each two consecutive terms n is the position of the
number
- The sum of first n terms of an Arithmetic sequence is calculate from
Sn = n/2[a + l], where a is the first term and l is the last term
* Now lets solve the problem
- We will use method (1)
- From the table the terms of the sequence are:
10 , 22 , 34 , 46 , 58 , 82 , 94 , ............., where 10 is the first term
∵ an = a1 + (n - 1) d ⇒ explicit formula
∵ a1 = 10 and a2 = 22
∵ d = a2 - a1
∴ d = 22 - 10 = 12
- The 100th term means the term of n = 100
∴ a100 = 10 + (100 - 1) 12
∴ a100 = 10 + 99 × 12 = 10 + 1188 = 1198
∴ The 100th term is 1198
- Lets find the sum of the first 100 terms of the sequence
∵ Sn = n/2[a1 + an]
∵ n = 100 , a = 10 , a100 = 1198
∴ S100 = 100/2[10 + 1198] = 50[1208] = 60400
* The sum of the first 100 terms is 60400
