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If in the expansion of (1 + x) n , the coefficients of three consecutive terms are 28, 56 and 70.Then find n and the position of terms of these coefficients.

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Let terms be nCr, nCr + 1, Cr + 2.

Now, After cross multiplying, we get 

n – r = 2 (r + 1) 

⇒ n = 3r + 2 

⇒ 4n = 12r + 8    (i)

After cross multiplying, we get 

4(n – r – 1) = 5(r + 2) 

⇒ 4 = 9r + 14 

Equating (i) and (ii), we get 

9r + 14 = 12r + 8 

⇒ 3r = 6 or r = 2 

Substituting the value of r in (i), we get

 4n = 12 × 2 + 8 or 4n = 32 orn = 8

Position of 3 consecutive terms are 3, 4 and 5. i.e., 3rd term, 4th term and 5th term, respectively.

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