Question

2. Show that the progression 16, 11, 6, 1,4,... is an AP.
Write down its first term and the common difference.​

Answer 1

Step-by-step explanation:

a =16

a2 = 11

a3 =6

a2-a1 = 11-16=-5

a3-a2 = 6-11 =-5

a3-a2 = a2-a

so d is same in both eqn hence this progression is in AP. 16,11,6,1,-4 .... is an Ap

d =a2-a =-5

a =16

Answer 2

Answer:

this progression is AP

Step-by-step explanation:

If= a2-a1=d=11-16=-5

=a3-a2=d=6-11=-5

If all time we get a common difference than the progression is AP.

common difference d= -5

and first term is a is 16.

an=a+{n-1}d

let a3=6 is the finite last term of the progression.

n=3

d=-5 and

a=a3-{3-1}{-5}

a=6-{-10}

a=6+10

a=16.

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