Question

Fourth term from the end of the sequence 27, 19, 11, ............., –125 is

Answer 1

Answer:

$\bigstar{\bold{Fourth\:term\:from\:end\:=\:-101}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• The A.P is 27,19,11.......-125

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Fourth term from the end of sequence

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ The fourth term from the end of sequence is given by

  The term from the end = (l -(n-1)d)

   Where l is the last term of the A.P and d is the common difference

→ In this AP common difference = Second term - First term

  d = 19-27 = -8

→ The last term of this AP is -125

→ To find the value of fourth term from end, substitute the data in above equation.

→ Fourth term from end =(-125-(4-1)-8)

  Fourth term from end = -125+24

  $\boxed{\bold{Fourth\:term\:from\:end\:=\:-101}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The common difference of an AP is the difference between its two consecutive terms.

→ It can be found out by two methods

• $d=a_2-a_1$

• $d=\frac{a_m-a_n}{m-n}$