Question

find the value of x if
$\sqrt[3]{4x - 5} - 5 = 0$
​

Answer 1

★Answer★:

(4x-5) ^1/3-5=0

→ (4x-5) ^1/3=5

→ (4x-5) =(5) ³

→ 4x-5=125

→ 4x=125+5

→ 4x=130

→ x=130/4

→ x=32.5

So the value of x is 32.5

Answer 2

Answer:

ANSWER

3, 15, 27, 39…….

a

n

=a+(n−1)d

a=3,d=15−3=12

n=54

a

54

=a+(n−1)d

=3+(54−1)×12

=3+53×12

=639

Term which is 132 more than its 54th term is – 639 +132 = 771.

a

n

=771

771=a+(n−1)d

771=3+(n−1)12

771=3+(n−1)12

64=n−1

n=65