Question

The two consecutive terms in the expansion of ${(3+2x)}^{74}$ whose coefficient are equal, is/are

• A. ${30}^{th}$ and ${31}^{st}$ terms
• B. ${29}^{th}$ and ${30}^{th}$ terms
• C. ${31}^{st}$ and ${32}^{nd}$ terms
• D. ${28}^{th}$ and ${29}^{th}$ terms

Answer 1

The correct option is A ${30}^{th}$ and ${31}^{st}$ terms

$T_{r+1}=T_{r+2}$
Considering, only the coefficients, we get
${}^n{C}_r{3}^{n-r}{2}^r={}^n{C}_{r+1}{3}^{n-r-1}{2}^{r+1}$
$3{}^n{C}_r=2{}^n{C}_{r+1}$
$3(r+1)=2(n-r)$
Now $n=74$
$3r+3=148-2r$
$5r=145$
$r=29$
Hence the terms are $T_{30}$ and $T_{31}$.