Binomial Theorem

Let 3 consecutive terms in the expansion of (1+x)23 be
Tr+1,Tr+2‌ and ‌Tr+3
Coefficient of Tr+1=‌23Cr
Coefficient of Tr+2=‌23Cr+1
Coefficient of Tr+3=‌23Cr+2
According to the question,
‌23Cr+1−‌23Cr‌=‌23Cr+2−‌23Cr+1
⇒
‌=‌

23!

(r+1)!(22−r)!

−‌

23!

r!(23−r)!

⇒‌‌‌

23!

(r+2)!(21−r)!

−‌

23!

(r+1)!(22−r)!

⇒
‌=‌

23−r

(r+1)!(23−r)!

−‌

(r+1)

(r+1)!(23−r)!

⇒‌‌
⇒‌‌(r+2)[(23−r)−(r+1)]
⇒‌‌(23−r)[(22−r)−(r+2)]
⇒
⇒(r+2)(22−2r)=(23−r)(20−2r)

⇒r2−84r+416=0
⇒(r−8)(r−13)=0
⇒r=8,13
Hence, 3 consecutive terms are either T9,T10 and T11 or T14,T15 and T16.