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Worked Examples · Example 11

Q.Solve for xx: 1+4+7+10+…+x=5901 + 4 + 7 + 10 + \ldots + x = 590

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Recognizing 1+4+7+⋯+x1+4+7+\cdots+x as an A.P. sum, solving Sn=590S_n=590 for nn gives n=20n=20, and then xx is the 20th term, 5858.

Sum of the first nn terms of an A.P.:

Sn=n2[2a+(n−1)d],an=a+(n−1)dS_n = \frac{n}{2}\big[2a+(n-1)d\big], \qquad a_n = a+(n-1)d

where aa is the first term, dd the common difference, and x=anx=a_n the last (unknown) term.

  1. The series 1,4,7,10,…,x1,4,7,10,\ldots,x is an A.P. with a=1a=1 and d=4−1=3d=4-1=3.
  2. Let there be nn terms up to and including xx. Then Sn=n2[2(1)+(n−1)(3)]=n2[3n−1]S_n = \dfrac{n}{2}\big[2(1)+(n-1)(3)\big] = \dfrac{n}{2}\big[3n-1\big].
  3. Set Sn=590S_n=590: n(3n−1)2=590⇒n(3n−1)=1180⇒3n2−n−1180=0\dfrac{n(3n-1)}{2} = 590 \Rightarrow n(3n-1) = 1180 \Rightarrow 3n^2-n-1180=0.
  4. Solve the quadratic: discriminant =(−1)2−4(3)(−1180)=1+14160=14161= (-1)^2-4(3)(-1180) = 1+14160 = 14161, and 14161=119\sqrt{14161}=119. …

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