The fifth term of an arithmetic sequence is 22 and the 15th term is 62. Find the 100th term and the sum of the first 60 terms

T15-T5 = 10d = 40

so, d = 4
Now you can find T1 and T60

Then the sum is 30(T1+T60)

To find the 100th term of the arithmetic sequence, we can first find the common difference (d) and the first term (a₁).

Since the fifth term is 22, we can use the formula for the nth term of an arithmetic sequence:
aₙ = a₁ + (n - 1) * d

Substituting the values we know:
22 = a₁ + (5 - 1) * d
22 = a₁ + 4d --- (Equation 1)

Also, the fifteenth term is 62:
62 = a₁ + (15 - 1) * d
62 = a₁ + 14d --- (Equation 2)

To solve this system of linear equations, we can subtract Equation 1 from Equation 2 in order to eliminate a₁:

62 - 22 = (a₁ + 14d) - (a₁ + 4d)
40 = 10d

Dividing both sides by 10:
d = 4

Now that we have the value of d (common difference), we can substitute it into Equation 1 to find a₁:

22 = a₁ + 4 * 4
22 = a₁ + 16

Subtracting 16 from both sides:
a₁ = 6

We now know that the first term (a₁) is 6 and the common difference (d) is 4.

To find the 100th term (a₁₀₀), we'll use the formula:
aₙ = a₁ + (n - 1) * d

Substituting the values we have:
a₁₀₀ = 6 + (100 - 1) * 4
a₁₀₀ = 6 + 99 * 4
a₁₀₀ = 6 + 396
a₁₀₀ = 402

Therefore, the 100th term of the arithmetic sequence is 402.

To find the sum of the first 60 terms, we can use the formula for the sum of an arithmetic series:

Sₙ = (n/2) * (a₁ + aₙ)

Substituting the values we have:
S₆₀ = (60/2) * (6 + a₆₀)

Since we already have a₁ and a₆₀ is the 60th term, we can calculate a₆₀ using the formula for the nth term:

a₆₀ = a₁ + (60 - 1) * d

Substituting the values we know:
a₆₀ = 6 + (60 - 1) * 4
a₆₀ = 6 + 59 * 4
a₆₀ = 6 + 236
a₆₀ = 242

Now, we can substitute a₆₀ into the formula for the sum of an arithmetic series:
S₆₀ = (60/2) * (6 + 242)
S₆₀ = 30 * 248
S₆₀ = 7440

Therefore, the sum of the first 60 terms of the arithmetic sequence is 7440.