QJEEM22Ma80f5: JEE Main 2022 Mathematics - Class 11 JEEM22Ma80f5 Solution: If n arithmetic means are inserted between a and 100 such that…

a, A₁, A₂ ........... A_n, 100

Let d be the common difference of above A.P. then

${{a + d} \over {100 - d}} = {1 \over 7}$

$\Rightarrow 7a + 8d = 100$ ...... (i)

and $a + n = 33$ ..... (ii)

and $100 = a + (n + 1)d$

$\Rightarrow 100 = a + (34 - a){{(100 - 7a)} \over 8}$

$\Rightarrow 800 = 8a + 7{a^2} - 338a + 3400$

$\Rightarrow 7{a^2} - 330a + 2600 = 0$

$\Rightarrow a = 10,\,{{260} \over 7},$ but $a \ne {{260} \over 7}$

$\therefore$ $n = 23$