Question

An AP consists of 50 term of which 3rd term is 12 and the last term is 106 find the 25 terms​

Answer 1

Answer:

Required Answer:-

Given question:

$\large{ \boxed{\rm{ {5}^{2x - 1} = \frac{1}{ {125}^{x - 3} } \: \: find \: x }}}$

Here,

First express 125 in terms of power of 5.

Nextly, equalise the bases i.e. 5

Compute the relation between powers of 5 in both sides of the equation.

Solving...

Expressing in terms of 5 both sides,

➛$\rm{5}^{2x - 1} = \dfrac{1}{ ({5}^{3}) {}^{x - 3} }$

When the power is negative, it means the value is in the denominator with the same magnitude power (exponent).

➛$\rm{ {5}^{2x - 1} = \dfrac{1}{ {5}^{3x - 9} } }$

➛$\rm{5}^{2x - 1} = {5}^{ - (3x - 9)}$

Now, the bases are same. We can directly do the computation upon the powers.

➛$\rm2x - 1 = - (3x - 9)$

➛$\rm{2x - 1 = - 3x + 9}$

➛$\rm{2x + 3x = 9 + 1}$

➛$\rm 5x = 10$

➛$\rm{\red{x = 2}}$

Hence:-

The required value of x, as asked in the question is 2 (Ans).

Answer 2

Answer:

$\bf{Explanation:}$

a+2d=12

a+49d=106

⇒47d=94

⇒d=2

a+4=12

⇒a=8

29th term : -

a+28d=8+56=64