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Worked Examples · Example 2

Q.In a sequence, the nnth term is an=2n2+5a_n = 2n^2+5. Show that it is not an A.P.

Sikkim CbseNCERTSubjective· 2mImportance★★★★★est
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✓ Free question

A sequence is an A.P. only if the difference between every pair of consecutive terms is constant; for an=2n2+5a_n=2n^2+5 this difference changes, so it is not an A.P.

A sequence a1,a2,a3,…a_1,a_2,a_3,\ldots is an Arithmetic Progression (A.P.) iff an+1−an=da_{n+1}-a_n = d (a constant) for every nn.

  1. Given an=2n2+5a_n = 2n^2+5.
  2. a1=2(1)2+5=2+5=7a_1 = 2(1)^2+5 = 2+5=7.
  3. a2=2(2)2+5=8+5=13a_2 = 2(2)^2+5 = 8+5=13.
  4. a3=2(3)2+5=18+5=23a_3 = 2(3)^2+5 = 18+5=23.
  5. First difference: a2−a1=13−7=6a_2-a_1 = 13-7=6.
  6. Second difference: a3−a2=23−13=10a_3-a_2=23-13=10.
  7. Since 6≠106 \ne 10, the difference between consecutive terms is not constant, violating the definition of an A.P.
✓Final answer

The sequence is not an A.P., because a2−a1=6a_2-a_1=6 while a3−a2=10a_3-a_2=10 — the common-difference condition fails.

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