Question

3.
12 term of the AP 5, 8-11, 14, .... is:​

Answer 1

Answer :

12th term = 38

Step-by-step explanation :

Arithmetic Progression :

• It is the sequence of numbers such that the difference between any two successive numbers is constant.

• In AP,

      a - first term

      d - common difference

      n - number of terms

       l - last term

     aₙ - nth term

     Sₙ - sum of n terms

• General form of AP,

      a , a+d , a+2d , a+3d , ..........

• Sum of n terms,

           $\boxed{\bf S_n=\frac{n}{2}[2a+(n-1)d]}$

           $\boxed{\bf S_n=\frac{n}{2}[a+l]}$

_____________________________

Given AP series,

5 , 8 , 11 , 14 , ....

First term, a = 5

common difference, d = 8 - 5 = 11 - 8 = 3

nth term is given by,

  aₙ = a + (n - 1)d

we have to find 12th term,

substitute n = 12 and the other values,

 a₁₂ = 5 + (12 - 1) (3)

      = 5 + 11(3)

      = 5 + 33

      = 38

∴ 12th term = 38