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.