AMC 10B 2022 (Problem 15)Let SnS_nSn be the sum of the first nnn terms of an arithmetic sequence that has a common difference of 222. The quotient S3nSn\dfrac{S_{3n}}{S_n}SnS3n does not depend on nnn. What is S20S_{20}S20?(A) 340\text{(A)}\;340(A)340(B) 360\text{(B)}\;360(B)360(C) 380\text{(C)}\;380(C)380(D) 400\text{(D)}\;400(D)400(E) 420\text{(E)}\;420(E)420Related TopicsCoreToolkit 25 — Arithmetic Sequence and SeriesHints (9)Hint 1an=a1+(n−1)d=a1+2(n−1)a_n = a_1 + (n-1)d = a_1 + 2(n-1)an=a1+(n−1)d=a1+2(n−1)Hint 2Sn=n2(a1+an)=n2(2a1+2(n−1))S_n = \frac{n}{2}(a_1+a_n) = \frac{n}{2}\big(2a_1+2(n-1)\big)Sn=2n(a1+an)=2n(2a1+2(n−1))Hint 3S3n=3n2(2a1+2(3n−1))S_{3n} = \frac{3n}{2}\big(2a_1+2(3n-1)\big)S3n=23n(2a1+2(3n−1))Hint 4Let a=a1a=a_1a=a1. S3nSn=3(2a+2(3n−1))2a+2(n−1)\frac{S_{3n}}{S_n} = \frac{3\big(2a+2(3n-1)\big)}{2a+2(n-1)}SnS3n=2a+2(n−1)3(2a+2(3n−1))Hint 5Since the ratio is independent of nnn, S3S1=S6S2\dfrac{S_3}{S_1}=\dfrac{S_6}{S_2}S1S3=S2S6.Hint 63(2a+4)2a=3(2a+10)2a+2\frac{3(2a+4)}{2a} = \frac{3(2a+10)}{2a+2}2a3(2a+4)=2a+23(2a+10)Hint 7a+2a=a+5a+1\frac{a+2}{a} = \frac{a+5}{a+1}aa+2=a+1a+5Hint 8a2+5a=a2+3a+2⇒2a=2⇒a=1a^2+5a = a^2+3a+2 \Rightarrow 2a=2 \Rightarrow a=1a2+5a=a2+3a+2⇒2a=2⇒a=1Hint 9S20=202(a1+a20)=10(1+(1+19⋅2))=400S_{20} = \frac{20}{2}(a_1+a_{20}) =10\Big(1 + \big(1+19\cdot2\big)\Big) =400S20=220(a1+a20)=10(1+(1+19⋅2))=400Final Answer(D) 400Related Problems (1)MATHCOUNTS 2026 State Target Round (Problem 5)