Answer:
The value of the sum is 1078
Explanation:
* Lets revise how to find the sum of the arithmetic series
- In the arithmetic series there is a constant difference between each
two consecutive terms
- Ex:
# 4 , 9 , 14 , 19 , 24 , .......... (+ 5)
# 25 , 15 , 5 , -5 , -15 , .......... (-10)
- So if the first term is a and the constant difference between each two
consecutive terms is d, then
U1 = a , U2 = a + d , U3 = a + 2d , U4 = a + 3d , ..........
- Then the nth term is Un = a + (n - 1)d
- The sum of n terms is Sn = n/2[a + L] , where L is the last term in
the series
* Lets solve the problem
∵ The arithmetic series is 7 + 11 + 15 + 19 + ......................... + 91
∴ The first term (a) is 7
∴ The last term is (L) 91
- Lets find the constant difference
∵ 11 - 7 = 4 and 15 - 11 = 4
∴ The constant difference (d) is 4
- Lets find the sum of the series from 7 to 91
∵ Sn = n/2[a + L]
∵ a = 7 and L = 91
∴ Sn = n/2[7 + 91]
- We need to know how many terms in the series
∵ L is the last term and equals 91 lets find its position in the series
∵ Un = a + (n - 1)d
∵ a = 7 , d = 4 and Un = 91
∴ 91 = 7 + (n - 1)(4) ⇒ subtract 7 from both sides
∴ 84 = (n - 1)(4) ⇒ divide both sides by 4
∴ 21 = n - 1 ⇒ add 1 to both sides
∴ n = 22
∴ The number of the terms in the series is 22
- Lets find the sum of the 22 terms (S22)
∴ S22 = 22/2[7 + 91]
∴S22 = 11[98] = 1078
* The value of the sum is 1078