Question

11th term from the last of the A.P. 10,12,14...50 is:
40
34
32
30

Answer 1

Answer :

11th term from the last of the given A.P. = 30

Step-by-step explanation :

$\underline{\bf Arithmetic \ Progression:}$

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

• General form of AP,

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

Given,

A.P. series : 10 , 12 , 14 , ..... , 50

To find,

11th term from the end = ?

Solution,

Let's reverse the A.P.

50, ....., 14 , 12 , 10

first term, a = 50

common difference, d = 10 - 12 = -2

we have to find the 11th term,

i.e., a₁₁ = ?

we know that,

nth term of an A.P., is given by,

$\boxed{\bf a_n=a+(n-1)d}$

a₁₁ = 50 + (11 - 1) (-2)

a₁₁ = 50 + 10(-2)

a₁₁ = 50 - 20

a₁₁ = 30

∴ 11th term from the end of the A.P., 10,12,14, ....,50 is 30