Question

2. Show that the sequence defined by an= 3n² - 5 is not an AP​

Answer 1

Answer:

Hi ,

We know that ,

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A list of numbers in which each term is

obtained by adding or subtracting a fixed

number to the preceeding terms , except

the first term is called an arithmetic

progression simply A.P.

The difference between any two successive

terms is the same throughout the series , this

is called common difference .

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According to the problem given ,

an = 3n^2 - 5 ----( 1 )

Put n = 1 in equation ( 1 ),

1 ) a1 = 3× 1^2 - 5 = 3 - 5 = -2

2 ) put n = 2 in ( 1 ) ,

a2 = 3 × 2^2 - 5 = 12 - 5 = 7

3 ) put n = 3 in ( 1 ) ,

a3 = 3 × 3^2 - 5 = 27 - 5 = 22

Now ,

a2 - a1 = 7 - ( - 2 ) = 7 + 2 = 9 ---( 2 )

a3 - a 2 = 22 - 7 = 15 -------( 2 )

From ( 2 ) and ( 3 ) ,

a2 - a1 is not equals to a3 - a2

Therefore,

an is not an A. P.

Step-by-step explanation:

Answer 2

Step-by-step explanation:

if an-a(n-1) is independent of n then it is A. P

or find the terms

a1=3*1-5=-2

a2=3*4-5=7

a3=3*9-5=22

so d= a2-a1=7-2=5

a3-a2=d

a3-a2=22-7=15

in both cases d is not equal so it won't form AP