SOLUTION: The first and last terms of an A.P are 1 and 121 respectively. Find (a) the number of terms in the A.P (b) the commons difference if the sum of its term is 671

Question 1209188: The first and last terms of an A.P are 1 and 121 respectively. Find
(a) the number of terms in the A.P
(b) the commons difference if the sum of its term is 671
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

n = number of terms
d = common difference
a₁ = first term = 1
a_n = n^th term = 121

S_n = sum of the first n terms of an arithmetic sequence
S_n = (n/2)*(a₁+a_n)
671 = (n/2)*(1+121)
671 = (n/2)*122
671 = 61n
n = 671/61
n = 11
There are 11 terms in this arithmetic sequence.

a_n = n^th term
a_n = a₁ + d*(n-1)
a₁₁ = 1 + d*(11-1)
121 = 1 + d*(11-1)
121 = 1 + 10d
10d = 121-1
10d = 120
d = 120/10
d = 12
The common difference is 12.

The arithmetic sequence is: 1, 13, 25, 37, 49, 61, 73, 85, 97, 109, 121
I used GeoGebra to generate the sequence quickly. You can also use a spreadsheet to generate the list.
The sum of these values is 671 to help confirm the answers are correct.
I don't recommend manually typing 1+13+25+37+49+61+73+85+97+109+121 into a calculator such as a TI83 since it would be tedious busy-work.

Answers:
(a) 11 terms
(b) common difference is 12

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