Algebra · real student question

An arithmetic sequence has u2 = 1 and u5 = 19. The number 103 is which term of the sequence?

Question

An arithmetic sequence (un)(u_n) has u2=1u_2=1 and u5=19u_5=19. The number 103103 is which term of the sequence?

A. 1717 B. 1818 C. 1919 D. 2020

Step-by-step solution

  1. Find dd from the two given terms. u5−u2=(5−2)d=3d⟹3d=19−1=18, d=6u_5-u_2=(5-2)d=3d\quad\Longrightarrow\quad 3d=19-1=18,\ d=6

  2. Recover the first term. u2=u1+d=1u_2=u_1+d=1, so u1=1−6=−5u_1=1-6=-5

  3. Set the general term equal to 103. −5+(n−1)⋅6=103-5+(n-1)\cdot 6=103

  4. Solve for nn. 6(n−1)=108⟹n−1=18⟹n=196(n-1)=108\quad\Longrightarrow\quad n-1=18\quad\Longrightarrow\quad n=19 The integer result is what makes 103103 genuinely a member of the sequence; a fractional nn would mean it is not.

  5. Check directly. u19=−5+18⋅6=−5+108=103u_{19}=-5+18\cdot 6=-5+108=103, so the answer is C.

Answer

n=19(u19=103)n=19\quad(u_{19}=103)

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