a+d = 5/8
a + 8d = 71/24
subtract
7d = 7/3
d = 1/3
then in a+d = 5/8
a = 5/8 - 1/3 = 7/24
term 265 = a + 264d
=7/24 + 264/3 = 2119/24
a + 8d = 71/24
subtract
7d = 7/3
d = 1/3
then in a+d = 5/8
a = 5/8 - 1/3 = 7/24
term 265 = a + 264d
=7/24 + 264/3 = 2119/24
Let's begin by finding the common difference (d):
We are given the second term, which is 5/8, and the ninth term, which is 71/24.
Using the formula for the nth term of an arithmetic sequence, we can write the equations:
a + (n - 1)d = t₁, where a is the first term (unknown), n is the position of the term, and t₁ is the given term.
For the second term, substituting n = 2 and t₁ = 5/8, we get:
a + (2 - 1)d = 5/8
a + d = 5/8 (Equation 1)
Similarly, for the ninth term, substituting n = 9 and t₁ = 71/24, we get:
a + (9 - 1)d = 71/24
a + 8d = 71/24 (Equation 2)
We now have a system of two equations (Equation 1 and Equation 2) with two variables (a and d). We can solve this system to find the values of a and d.
To solve this system, we'll multiply Equation 1 by 8 and Equation 2 by 3 to eliminate the fractions:
8(a + d) = 8(5/8)
24(a + 8d) = 24(71/24)
This simplifies to:
8a + 8d = 5
24a + 192d = 71
Let's use elimination or substitution to solve this system of equations:
Multiply Equation 1 by 3:
3(a + d) = 3(5/8)
3a + 3d = 15/8 (Equation 3)
Subtract Equation 3 from Equation 2:
(24a + 192d) - (3a + 3d) = (71) - (15/8)
24a + 192d - 3a - 3d = 568/8 - 15/8
21a + 189d = 553/8
Now, we have a new equation:
21a + 189d = 553/8 (Equation 4)
Let's solve this new equation (Equation 4) with Equation 3 to eliminate one variable:
Multiply Equation 3 by 7:
7(3a + 3d) = 7(15/8)
21a + 21d = 105/8 (Equation 5)
Subtract Equation 5 from Equation 4:
(21a + 189d) - (21a + 21d) = (553/8) - (105/8)
21a + 189d - 21a - 21d = 448/8
168d = 448/8
d = (448/8) / 168
d = (448/8) * (1/168)
d = 28/168
d = 1/6
Now that we have found the common difference (d = 1/6), we can find the first term (a) using Equation 1:
a + d = 5/8
a + 1/6 = 5/8
a = 5/8 - 1/6
a = (15/24) - (4/24)
a = 11/24
Thus, the first term (a) is 11/24 and the common difference (d) is 1/6.
Now, we can use the formula for the nth term of an arithmetic sequence to find the 265th term:
tₙ = a + (n - 1)d
Substituting the known values:
tₙ = (11/24) + (265 - 1)(1/6)
tₙ = (11/24) + (264/6)
tₙ = (11/24) + (44/24)
tₙ = 55/24
Therefore, the 265th term of the arithmetic sequence is 55/24.