Question

Find the mth term of an Arithmetic progression whose 12th
term exceeds the 5th term by 14 and the sum of both terms
is 36.

Answer 1

Answer:

the answer is mth term= 1+2m

Answer 2

★Question :

Find the mth term of an Arithmetic progression whose 12th  term exceeds the 5th term by 14 and the sum of both terms  is 36.

★Answer :

mth term = 1 + 2m

★Given :

• 12th  term exceeds the 5th term by 14
• the sum of both terms  is 36

★To find :

mth term

★Solution :

   In an AP, nth term is given by,

       $\boxed{a_n=a+(n-1)d}$

where

 a is the first term

 d is the common difference

 aₙ is the nth term

 

⇒ 12th term,

  $a_{12}=a+(12-1)d \\ a_{12}=a+11d$

⇒ 5th term,

  $a_5=a+(5-1)d \\ a_5=a+4d$

>>> 12th  term exceeds the 5th term by 14

       $a_{12}=a_5+14 \\ a+11d=a+4d+14\\11d=4d+14\\11d-4d=14\\7d=14\\d=14/7\\d=2$

$\underline{\underline{\bf common \ difference, d = 2}}$

>>> The sum of both terms = 36

      $a_{12}+a_5=36 \\a+11d+a+4d=36\\2a+15d=36\\2a+15(2)=36\\2a+30=36\\2a=36-30\\2a=6\\a=6/2\\a=3$

$\underline{\underline{\bf first \ term, a = 3}}$

=> Now, to find the mth term ;

          $a_m=a+(m-1)d \\\\ a_m=3+(m-1)2 \\\\ a_m=3+2m-2\\\\a_m=1+2m$

∴ mth term = 1+2m