Question

Which term of an Arithmetic Progression 6, 4, 2, 0, −2, .... is -120?​

Answer 1

An = -120

d = A2-A1 = 4-6 = -2

A1 = 6

An = A1 +(n-1)d

-120 = 6 +(n-1)*(-2)

-126 = -2n + 2

2n = 128

n = 64 th term

Answer 2

64th term of the arithmetic progression 6, 4, 2, 0, - 2 . . . . is - 120

Given :

The arithmetic progression 6, 4, 2, 0, - 2 . . . .

To find :

Which term of the AP 6, 4, 2, 0, - 2 . . . is - 120

Concept :

If in an arithmetic progression

First term = a

Common difference = d

Then nth term of the AP

= a + ( n - 1 )d

Solution :

Step 1 of 3 :

Write down the given progression

Here the given arithmetic progression is

6, 4, 2, 0, - 2 . . . .

Step 2 of 3 :

Write down first term and common difference

First term = a = 6

Common Difference = d = 4 - 6 = - 2

Step 3 of 3 :

Find which term of the AP 6, 4, 2, 0, - 2 . . . is - 120

Let nth term of the AP 6, 4, 2, 0, - 2 . . . . is - 120

Then nth term of the AP = - 120

a + ( n - 1 )d = - 120

⇒ 6 + ( n - 1 ) × ( - 2 ) = - 120

⇒ 6 - 2n + 2 = - 120

⇒ - 2n + 8 = - 120

⇒ - 2n = - 120 - 8

⇒ - 2n = - 128

⇒ n = 64

Hence 64th term of the arithmetic progression 6, 4, 2, 0, - 2 . . . . is - 120

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