AMC 10A/12A 2026 (Problem 5/4)Let a1,a2,a3,…a_1,a_2,a_3,\ldotsa1,a2,a3,… be a sequence with a1=94a_1=94a1=94, a2=190a_2=190a2=190, and ana_nan equal to the arithmetic mean of the previous n−1n-1n−1 terms for n≥3n\ge 3n≥3. What is the value of a10a_{10}a10?(A) 47\text{(A)}\;47(A)47(B) 95\text{(B)}\;95(B)95(C) 142\text{(C)}\;142(C)142(D) 284\text{(D)}\;284(D)284(E) 564\text{(E)}\;564(E)564Related TopicsCoreTelescoping Approaches to Solving RecurrencesCheck AnswerYour answer:ABCDECheckHints (4)Hint 1an=a1+a2+⋯+an−1n−1a_n=\frac{a_1+a_2+\cdots+a_{n-1}}{n-1}an=n−1a1+a2+⋯+an−1Hint 2(n−1)an=a1+a2+⋯+an−1(n-1)a_n=a_1+a_2+\cdots+a_{n-1}(n−1)an=a1+a2+⋯+an−1nan+1=a1+a2+⋯+an−1+anna_{n+1}=a_1+a_2+\cdots+a_{n-1}+a_nnan+1=a1+a2+⋯+an−1+annan+1−(n−1)an=anna_{n+1}-(n-1)a_n=a_nnan+1−(n−1)an=anHint 3nan+1=nanna_{n+1}=na_nnan+1=nanan+1=ana_{n+1}=a_nan+1=anHint 4a10=a9=⋯=a3=a1+a22=94+1902a_{10}=a_9=\cdots=a_3=\frac{a_1+a_2}{2}=\frac{94+190}{2}a10=a9=⋯=a3=2a1+a2=294+190Answer=a10=2842=142\text{Answer}=a_{10}=\frac{284}{2}=142Answer=a10=2284=142Final Answer(C) 142Related Problems (2)AMC 12A 2023 (Problem 20)2024 AMC 12A Problem 21