The fourth term of an AP is 11 and the eighth term exceeds twice the fourth term by five.Find the AP and the sum of first 50 terms.

Answer 1

a1 = the initial term of an arithmetic progression

d = the common difference of successive members

nth term of the sequence:

an = a1 + ( n - 1 ) ∙ d

In this case:

a4 = 11

a4 = a1 + ( 4 - 1 ) ∙ d

a4 = a1 + 3 d

11 = a1 + 3 d

a8 = 2 a4 + 5

a8 = 2 ∙ 11 + 5 = 22 + 5 = 27

a8 = a1 + ( 8 - 1 ) ∙ d

a8 = a1 + 7 d

a8 - a 4 = a1 + 7 d - ( a1 + 3 d ) =

a1 + 7 d - a1 - 3 d = 7 d - 3 d = 4 d

a8 - a 4 = 4 d = 27 - 11 = 16

4 d = 16

d = 16 / 4 = 4

d = 4

a4 = a1 + 3 d

11 = a1 + 3 ∙ 4

11 = a1 + 12

11 - 12 = a1

- 1 = a1

a1 = - 1

The sum of the n members of a arithmetic progression:

Sn = ( n / 2 ) ∙ [ 2 a1+ ( n - 1 ) ∙ d ]

In this case:

S50 = ( 50 / 2 ) ∙ [ 2 ∙ ( - 1 ) + ( 50 - 1 ) ∙ 4 ]

S50 = 25 ∙ ( - 2 + 49 ∙ 4 )

S50 = 25 ∙ ( - 2 + 196 )

S50 = 25 ∙ 194

S50 = 4850