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

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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.

i think it is 70

step-by-step explanation:

the anwser is d) 75°

step-by-step explanation: