If the coefficients of $r^{\text {th }},(r+1)^{\text {th }}$ and $(r+2)^{\text {th }}$ terms in the…
If the coefficients of $r^{\text {th }},(r+1)^{\text {th }}$ and $(r+2)^{\text {th }}$ terms in the expansion of $(1+x) n$ are in the ratio of $4: 15: 42$, then $n-r=$
$18$
$15$
$14$
$17$
Solution
General term of $(1+x)^{\mathrm{n}}$ is $\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} x^{\mathrm{r}}$
Coefficient of $r^{\text {th }}$ term $={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}$
Coefficient of $(r+1)^{\text {th }}$ term $={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$
Coefficient of $(r+2)^{\text {th }}$ term $={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}$
${ }^n C_{r-1}:{ }^n C_r:{ }^n C_{r+1}=4: 15: 42$
$\frac{{ }^n \mathrm{C}_{r-1}}{{ }^n \mathrm{C}_r}=\frac{4}{15} \Rightarrow 19 r-4 n=4$ ...(i)
And, $\frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{15}{42} \Rightarrow 19 r-5 n=-14$ ...(ii)
Subtracting (ii) from (i) $n=18$ and $r=4 \therefore n-r=14$