The first three terms of an arithmetic sequence are 36, 40, 44, ...

Question

7a. The first three terms of an arithmetic sequence are 36, 40, 44, ... 

(i) Write down the value of d. 

(ii) Find u₈ .

7b. 

(i) Show that S_n = 2n² + 34n. 

(ii) Hence, write down the value of S_14. 

Answer 1

Since we get from one term to the next by adding 4, d = 4.

 

If by u₈ you mean the 8th term, we use the formula

u_n = u₁ + (n-1)d to obtain u₈  = 36 + 4(8-1) = 64.

 

S_n = sum of first n terms = (n/2)[u₁ + u_n]

 

    = (n/2)(36 + 36 + 4(n-1))

 

    = (n/2)(68 + 4n) = 2n² + 34n

 

S₁₄ = 2(14)² + 34(14) = 868