QJEEAdv15Mad75b: JEE Advanced 2015 Mathematics - Class 11 JEEAdv15Mad75b Solution: Suppose that all the terms of an arithmetic progression (A.P)…

${{{S_7}} \over {{S_{11}}}} = {6 \over {11}}$ ...... (1)

$130 \le {t_7} \le 140$ ........ (2)

$\Rightarrow {{{7 \over 2}[2a + 6d]} \over {{{11} \over 2}[2a + 10d]}} = {6 \over {11}}$

$\Rightarrow {{a + 3d} \over {a + 5d}} = {6 \over 7}$ ....... (3)

$\Rightarrow {{{t_4}} \over {{4_6}}} = {6 \over 7}$

Let ${t_4} = 6k$ , ${t_6} = 7k$ ;

$2d = k \Rightarrow d = k/2$ and $a + 3d = 6k$

$\Rightarrow a = 6k - 3k/2 = 9k/2$

Hence, $130 \le {t_7} \le 140$ .

$\Rightarrow 130 \le {{9k} \over 2} + 3k \le 140$

$\Rightarrow 130 \le {{15k} \over 2} \le 140$

$\Rightarrow {{52} \over 3} \le k \in {{56} \over 3}$

Since, $k \in N \Rightarrow k = 18$ .

$\Rightarrow d = {k \over 2} = {{18} \over 2} = 9$