The sum of the 3rd term and 4th term of an arithmetic sequence is 18. The 7th term exceeds the 5th term by 14. Find the first term and common difference.

The first term and the common difference of the sequence are -8.5 and 7 respectively.

The sum of the 3rd term and 4th term of an arithmetic sequence is 18.

Using Arithmetic progression formula,

aₙ = a + (n- 1)d

where

n = nth term

a = first term

d = common difference

a₃ = a + 2d

a₄ = a + 3d

Therefore,

a + 3d + (a + 2d) = 18

2a + 5d = 18

The 7th term exceeds the 5th term by 14. Therefore,

a₅ = a + 4d

a₆ = a + 6d

a + 6d - (a + 4d) = 14

a + 6d - a - 4d  = 14

2d = 14

d = 14 / 2

d = 7

2a + 5d = 18

2a + 5(7) = 18

2a + 35 = 18

2a  = 18 - 35

2a = -17

a = -17 / 2

a = -8.5

learn more on arithmetic progression here: link

First term = -8.5.

Common difference = 7.

Step-by-step explanation:

The third term = a1 + 2d where a1 = first term and d is the common difference so we have

a1 + 2d + a1 + 3d = 18

2a1 + 5d = 18(1)

Also:

7th term - 5th term = 14 giving

a1 + 6d - ( a1 + 4d) = 14

2d = 14

d = 7.

So the common difference is 7 and substituting for d in (1) we have:

2a1 + 5*7 = 18

2a1 =  18-35 = -17

a1 = -8.5.

answer: 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

step-by-step explanation:

it is contrapositive

i could not see the question too well so check yourself again too . this is because i want you to get it correct.