Answer:
(a)11
(b)12
Step-by-step explanation:
The first term, a = 1
The last term, l=121
Sum of the series, [tex]S_n=671[/tex]
Given an arithmetic series where the first and last term is known, its sum is calculated using the formula:
[tex]S_n=\dfrac{n}{2}(a+l)[/tex]
Substituting the given values, we have:
[tex]671=\dfrac{n}{2}(1+121)\\671=\dfrac{n}{2} \times 122\\671=61n\\$Divide both sides by 61\\n=11[/tex]
(a)There are 11 terms in the arithmetic progression.
(b)We know that the 11th term is 121
The nth term of an arithmetic progression is derived using the formula:
[tex]a_n=a+(n-1)d[/tex]
[tex]a_{11}=121\\a=1\\n=11[/tex]
Therefore:
121=1+(11-1)d
121-1=10d
120=10d
d=12
The common difference between them is 12.
