Question

find the value of n if 2n+1,n^2+n+1,3n^2-3n+3 are consecutive terms of an AP​

Answer 1

Answer:

if a,b,c are in AP

b-a=c-b

2b = a+c

2(n²+n+1) = 2n+1+3n²-3n+3

2n²+2n+2 = 3n²-n+4

n²-3n+2 = 0

n²-n-2n+2 = 0

n(n-1)-2(n-1) = 0

(n-1)(n-2)=0

n=1,n=2

if n=1 : numbers are 3,3,3

if n=2: numbers are 5,7,9

Step-by-step explanation:

Answer 2

Answer:Brainly.in

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KarNage

Secondary School Math 5+3 pts

Find the value of n if 2n+1,n^2+n+1,3n^2-3n+3 are consecutive terms of an AP​

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Answers

KarNage

KarNageAmbitious

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TejaswisubrahmanyamExpert

Answer:

if a,b,c are in AP

b-a=c-b

2b = a+c

2(n²+n+1) = 2n+1+3n²-3n+3

2n²+2n+2 = 3n²-n+4

n²-3n+2 = 0

n²-n-2n+2 = 0

n(n-1)-2(n-1) = 0

(n-1)(n-2)=0

n=1,n=2

if n=1 : numbers are 3,3,3

if n=2: numbers are 5,7,9