Question

Difficulty: MediumArithmetic and Geometric Sequences and Series

In an arithmetic sequence, the sum of the first 44 terms is 2828 and the sum of the first 88 terms is 8888. What is the 10th10\text{th} term of this sequence?

  1. A
    1818
  2. B
    2020
  3. 2222Answer
  4. D
    2424
  5. E
    2626

Answer

The 10th term of the sequence is 22.
Using the arithmetic series sum formula Sn=n2(2a1+(n1)d)S_n = \frac{n}{2}(2a_1 + (n-1)d), the conditions yield two linear equations: 2a1+3d=142a_1 + 3d = 14 and 2a1+7d=222a_1 + 7d = 22. Subtracting these equations gives 4d=84d = 8, so d=2d = 2. Substituting d=2d = 2 back into 2a1+3(2)=142a_1 + 3(2) = 14 yields a1=4a_1 = 4. The 10th term is then calculated as a1+9d=4+9(2)=22a_1 + 9d = 4 + 9(2) = 22.

Step-by-Step Solution

1
Express the given sums using the arithmetic series sum formula Sn=n2(2a1+(n1)d)S_n = \frac{n}{2}(2a_1 + (n-1)d).
For S4=28S_4 = 28: 42(2a1+3d)=28    2a1+3d=14\frac{4}{2}(2a_1 + 3d) = 28 \implies 2a_1 + 3d = 14.
For S8=88S_8 = 88: 82(2a1+7d)=88    2a1+7d=22\frac{8}{2}(2a_1 + 7d) = 88 \implies 2a_1 + 7d = 22.
Setting up linear equations in terms of the first term a1a_1 and common difference dd allows us to solve for both sequence parameters.
2
Subtract the first equation from the second equation to solve for dd.
(2a1+7d)(2a1+3d)=2214    4d=8    d=2(2a_1 + 7d) - (2a_1 + 3d) = 22 - 14 \implies 4d = 8 \implies d = 2.
Eliminating a1a_1 isolates the common difference dd.
3
Substitute d=2d = 2 back into the first equation to solve for a1a_1.
2a1+3(2)=14    2a1+6=14    2a1=8    a1=42a_1 + 3(2) = 14 \implies 2a_1 + 6 = 14 \implies 2a_1 = 8 \implies a_1 = 4.
Finding a1a_1 completes the essential parameters of the sequence.
4
Calculate the 10th term using the formula an=a1+(n1)da_n = a_1 + (n-1)d.
a10=4+(101)(2)=4+18=22a_{10} = 4 + (10 - 1)(2) = 4 + 18 = 22.
Evaluating the formula at n=10n = 10 provides the target term.

Key Concept

Arithmetic sequence term and series sum formulas.
Estimated Time:2m 0s
Rate this question