JEE MAIN - Mathematics (2005 - No. 43)
If the coefficients of rth, (r+1)th, and (r + 2)th terms in the binomial expansion of $${{\rm{(1 + y )}}^m}$$ are in A.P., then m and r satisfy the equation
$${m^2} - m(4r - 1) + 4\,{r^2} - 2 = 0$$
$${m^2} - m(4r + 1) + 4\,{r^2} + 2 = 0$$
$${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$$
$${m^2} - m(4r - 1) + 4\,{r^2} + 2 = 0$$
Explanation
Let r = 2
$$\therefore$$ 2nd, 3rd and 4th terms are in AP.
2nd term = T2 = $${}^m{C_1}.y$$
Coefficient of T2 = $${}^m{C_1}$$
3rd term = T3 = $${}^m{C_2}.{y^2}$$
Coefficient of T3 = $${}^m{C_2}$$
4th term = T4 = $${}^m{C_3}.{y^3}$$
Coefficient of T2 = $${}^m{C_3}$$
$$\therefore$$ 2.$${}^m{C_2}$$ = $${}^m{C_1}$$ + $${}^m{C_3}$$
$$ \Rightarrow $$ $$2.{{m\left( {m - 1} \right)} \over {1.2}}$$ = $${m \over 1}$$ + $${{m\left( {m - 1} \right)\left( {m - 2} \right)} \over {1.2.3}}$$
$$ \Rightarrow $$ 6m2 - 6m = 6m +m(m2 - 3m + 2)
$$ \Rightarrow $$ 6m2 - 6m = 6m + m3 - 3m2 + 2m
$$ \Rightarrow $$ 6m - 6 = 6 + m2 - 3m + 2
$$ \Rightarrow $$ m2 - 9m + 14 = 0
Now put r = 2 at each option and find answer.
In option C, $${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$$ putting r = 2 we get
m2 - 9m + 14 = 0. So Option C is correct.
$$\therefore$$ 2nd, 3rd and 4th terms are in AP.
2nd term = T2 = $${}^m{C_1}.y$$
Coefficient of T2 = $${}^m{C_1}$$
3rd term = T3 = $${}^m{C_2}.{y^2}$$
Coefficient of T3 = $${}^m{C_2}$$
4th term = T4 = $${}^m{C_3}.{y^3}$$
Coefficient of T2 = $${}^m{C_3}$$
$$\therefore$$ 2.$${}^m{C_2}$$ = $${}^m{C_1}$$ + $${}^m{C_3}$$
$$ \Rightarrow $$ $$2.{{m\left( {m - 1} \right)} \over {1.2}}$$ = $${m \over 1}$$ + $${{m\left( {m - 1} \right)\left( {m - 2} \right)} \over {1.2.3}}$$
$$ \Rightarrow $$ 6m2 - 6m = 6m +m(m2 - 3m + 2)
$$ \Rightarrow $$ 6m2 - 6m = 6m + m3 - 3m2 + 2m
$$ \Rightarrow $$ 6m - 6 = 6 + m2 - 3m + 2
$$ \Rightarrow $$ m2 - 9m + 14 = 0
Now put r = 2 at each option and find answer.
In option C, $${m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0$$ putting r = 2 we get
m2 - 9m + 14 = 0. So Option C is correct.
Comments (0)
