Question

Find the value of the middle term of the A.P -6,-2,2,-.....,58. *​

Answer 1

Answer:

$\bigstar{\bold{Middle\:term=26}}$

Step-by-step explanation:

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

• A.P is -6, -2 , 2 , .....58

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

• The middle term of the A.P

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

→ First we have to find the number of terms(n) of the A.P

→ We know that last term of an A.P is given by,

   $a_n=a_1+(n-1)\times d$

  where a₁ = -6, $a_n$ = 58, d = -2 + 6 = 4

→ Substituting the datas we get the value for

   58 = -6 + (n - 1) × 4

   58 = -6 + 4n - 4

   58 = -10 + 4n

   4n = 58 + 10

   4n = 68

     n = 68/4

     n = 17

→ Hence the number of terms in the given A.P is 17

→ Since the number of terms is odd, the middle term would be (n+1)/2 th term

→ Hence the middle term here would be (17 + 1)/2 th term = 18/2 th term = 9th term.

→ Now we have to fid the ninth term of the A.P

→ Ninth term of the A.P is given by

  a₉ = a₁ + 8d

→ Substitute the datas,

  a₉ = -6 + 8 × 4

  a₉ = -6 + 32

  a₉ = 26

→ Hence the middle term of the A.P is 26

$\boxed{\bold{Middle\:term=26}}$

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

→ The nth term of an A.P is given by

   $a_n=a_1+(n-1)\times d$

→ Sum of n terms of an A.P is given by

  $S_n=\dfrac{n}{2}(a_1+a_n)$

 $S_n=\dfrac{n}{2}(2a_1+(n-1)\times d)$