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Worked Examples · Example 1

Q.Find the first three terms of the arithmetic progressions whose nnth term is given. Also find the common difference and the 20th term in each case.

(i) an=2n+5a_n = 2n+5
(ii) an=3−4na_n = 3-4n
(iii) an=n−34a_n = \dfrac{n-3}{4}
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Substitute n=1,2,3n=1,2,3 into each formula for the first three terms, take the difference of consecutive terms for dd, and substitute n=20n=20 for the 20th term.

For an A.P. with nnth term ana_n: first three terms are a1,a2,a3a_1,a_2,a_3; common difference d=a2−a1=a3−a2d=a_2-a_1=a_3-a_2; the 20th term is a20a_{20} found by substituting n=20n=20.

(i) an=2n+5a_n = 2n+5

  1. a1=2(1)+5=7a_1=2(1)+5=7, a2=2(2)+5=9a_2=2(2)+5=9, a3=2(3)+5=11a_3=2(3)+5=11.
  2. d=a2−a1=9−7=2d=a_2-a_1=9-7=2 (check: a3−a2=11−9=2a_3-a_2=11-9=2 ✓).
  3. a20=2(20)+5=40+5=45a_{20}=2(20)+5=40+5=45.

(ii) an=3−4na_n = 3-4n

4. a1=3−4(1)=−1a_1=3-4(1)=-1, a2=3−4(2)=−5a_2=3-4(2)=-5, a3=3−4(3)=−9a_3=3-4(3)=-9.

5. d=a2−a1=−5−(−1)=−4d=a_2-a_1=-5-(-1)=-4 (check: a3−a2=−9−(−5)=−4a_3-a_2=-9-(-5)=-4 ✓).

6. a20=3−4(20)=3−80=−77a_{20}=3-4(20)=3-80=-77.

(iii) an=n−34a_n=\dfrac{n-3}{4}

7. a1=1−34=−24=−12a_1=\dfrac{1-3}{4}=\dfrac{-2}{4}=-\dfrac12, a2=2−34=−14a_2=\dfrac{2-3}{4}=-\dfrac14, a3=3−34=0a_3=\dfrac{3-3}{4}=0. …

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