Arithmetic progression

Question

The third term of an arithmetic progression is 13 and the ninth term is 61. Find the first term and the common difference of the progression.

Answer 1

Hello Octavia,

The general equation for an arithmetic sequence is: a_n = a₁ + d*(n-1) where a₁ is the 1st term, a_n is the n-th term, and d is the common difference.

"The third term is 13" means that you replace an by 13 and n by 3 --> 13 = a₁ + d * (3 -1) ---> 13 = a₁ + 2d

"The ninth term is 61" means that you replace an by 61 and n by 9 --> 61 = a₁ + d *(9-1) ---> 61 = a₁ + 8d

Now you have two equations and two unknowns. You can eliminate a₁ by subtracting both equations:

13-61 = 2d - 8d ---> -48 = -6d -----> d = 8

Plug 8 in either equation: 13 = a₁ + 2*8 ---> 13 = a₁ + 16 ---> a₁ = -3



Checking our answers: a₃ = -3 + 8*(3-1) = -3 + 16 = 13 (YES!); a₉ = -3 + 8*(9-1) = -3+64 = 61 (YES!)

Therefore: the first term is -3 and the common difference is 8.