If the sum of the first 20 terms of the series log (71 / 2) x+ log (71 / 3) x+ log (71/4) x+ ldots is 460, then x is equal to :

Question

If the sum of the first 20 terms of the series log (71 / 2) x+ log (71 / 3) x+ log (71/4) x+ ldots is 460, then x is equal to : (A) 746 / 21 (B) 71 / 2 (C) e 2 (D) 72

Tardigrade Admin · Accepted Answer

460= log 7 x ⋅(2+3+4+ ldots . .+20+21) ⇒ 460= log 7 x ⋅((21 × 22/2)-1) ⇒ 460=230 ⋅ log 7 x ⇒ log 7 x=2 ⇒ x=49

Q. If the sum of the first $20$ terms of the series $lo g_{(7^{1/2})} x + lo g_{(7^{1/3})} x + lo g_{(7^{1/4})} x + \dots$ is $460,$ then $x$ is equal to :

$7^{46/21}$

$7^{1/2}$

$e^{2}$

$7^{2}$

$460 = lo g_{7} x \cdot (2 + 3 + 4 + \dots .. + 20 + 21)$
$\Rightarrow 460 = lo g_{7} x \cdot (\frac{21 \times 22}{2} - 1)$
$\Rightarrow 460 = 230 \cdot lo g_{7} x$
$\Rightarrow lo g_{7} x = 2$
$\Rightarrow x = 49$

Q. If the sum of the first 20 terms of the series log(71/2)​x+log(71/3)​x+log(71/4)​x+… is 460, then x is equal to :

746/21

71/2

e2

72