Find the 5th term of the arithmetic sequence whose 3rd term is 35 and whose 10th term is 77?

Answer 1

The 5th term is 47.

Answer 2

T(3)=35=>T(1)+(3-1)d=35---------(1)

T(10)=77=>T(1)+(10-1)d=77-------(2)

(2)-(1)=>

7d=42=>d=6 (the common difference)

T(1)=35-2(6)=23

Thus

T(5)=23+(5-1)6=47

Answer 3

The 5th term of the arithmetic sequence 

whose 3rd term is 35 

and whose 10th term is 77:

47

Answer 4

For an arithmetic sequence:

a₁

a₂ = a₁ + d ← where d is the common difference

a₃ = a₂ + d = a₁ + 2d

a₄ = a₃ + d = a₁ + 3d

a₅ = a₄ + d = a₁ + 4d

…and you can generalize by writing

a(n) = a₁ + (n - 1).d

a(n) = a₁ + (n - 1).d → when: n = 3

a₃ = a₁ + 2d → given that it's 35

a₁ + 2d = 35

a₁ = 35 - 2d

a(n) = a₁ + (n - 1).d → when: n = 10

a₁₀ = a₁ + 9d → given that it's 77

a₁ + 9d = 77 → recall the previous result: a₁ = 35 - 2d

35 - 2d + 9d = 77

7d = 42

d = 6

Recall:

a₁ = 35 - 2d → we've just seen that: d = 6

a₁ = 35 - 12

a₁ = 23

a₅ = a₁ + 4d

a₅ = 23 + (4 * 6)

a₅ = 47

Answer 5

t[3] = 35 = m + (3 - 1) * d = m + 2d

t[10] = 77 = m + (10 - 1) * d = m + 9d

77 - 35 = m + 9d - (m + 2d)

42 = m - m + 9d - 2d

42 = 7d

6 = d

35 = m + 2d

35 = m + 2 * 6

35 = m + 12

23 = m

t[n] = m + (n - 1) * d

t[n] = 23 + (n - 1) * 6

t[n] = 23 + 6n - 6

t[n] = 17 + 6n

t[5] = 17 + 6 * 5 = 17 + 30 = 47