Consider set $A = \{T (r + 1) =^{n} C_{r} (3)^{n - r} (5 x)^{r}; r = 0, 1, 2, 3, \dots, n\}$ . Match the following lists:
List I List II
a. For x=1, if T(12) is the greatest then the value of n can be p. 10
b. For n=25, if T(21) is the greatest then the value of 5x can be q. 18
c. For n=20, if T(10)=T(11) then the value of 33x can be r. 17
d. For x=2, if T(8) is the second largest then the value of n is s. 12

Moderate

Solution

a. If T(r + 1) is greatest, then $\frac{n}{1 + |\frac{a}{bx}|} \leq r \leq \frac{n + 1}{1 + |\frac{a}{bx}|}$ for (a + bx)ⁿ.
For x = 1 and r = 11, $\frac{n}{1 + \frac{3}{5}} \leq 11 \leq \frac{n + 1}{1 + \frac{3}{5}}$
$\begin{array}{l} \Rightarrow 5 n \leq 88 \leq 5 (n + 1) \\ \Rightarrow n = 17 \end{array}$
b. If T(r+ 1) is greatest, then $(\frac{r}{n - r + 1}) \frac{a}{b} < x < (\frac{r + 1}{n - r}) \frac{a}{b}$ for (a + bx)ⁿ.
For n = 25 and r = 20, $2 < x < \frac{63}{25}$
c. we have $\frac{T_{r + 1}}{T_{r}} = \frac{n - r + 1}{r} \cdot \frac{bx}{a} for (a + bx)^{n}$
For $n = 20 and r = 10, \frac{11}{10} \cdot \frac{5 x}{3} = 1 or x = 6 / 11$
d. We have $\frac{n}{1 + |\frac{a}{bx}|} \leq r \leq \frac{n + 1}{1 + |\frac{a}{bx}|} for (a + bx)^{n}$
For x = 2 and r = 8 (as T(9) will be largest),
$\begin{array}{l} \frac{n}{1 + \frac{3}{10}} \leq 8 \leq \frac{n + 1}{1 + \frac{3}{10}} \\ \Rightarrow 10 n \leq 104 \leq 10 (n + 1) \\ \Rightarrow n = 10 \end{array}$

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List I	List II
a. For x=1, if T(12) is the greatest then the value of n can be	p. 10
b. For n=25, if T(21) is the greatest then the value of 5x can be	q. 18
c. For n=20, if T(10)=T(11) then the value of 33x can be	r. 17
d. For x=2, if T(8) is the second largest then the value of n is	s. 12