Question

If the numbers n – 2, 4n – 1 and 5n + 2 are in AP, find the value of n

Answer 1

The terms are n - 2, 4n -1 and 5n + 2.
We know that if a , b and c are in A.P. then $\frac{a+c}{2}$=b
i.e. average of a and c is equal to b.
Therefore, by the condition:
$\frac{(n-2)+(5n+2)}{2}$=4n - 1.
⇒$\frac{n-2+5n+2}{2}$=4n -1
⇒$\frac{6n}{2}$=4n-1
⇒3n=4n-1
⇒4n-3n=1
⇒n=1
Ans:The value of n is 1 and the numbers are 1-2 = -1 , 4*1-1=3 and 5*1+2 = 7 respectively as we see that -1+4=3 and 3+4=7.
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Answer 2

Step-by-step explanation:

The terms are n - 2, 4n -1 and 5n + 2.

We know that if a , b and c are in A.P. then \frac{a+c}{2}

2

a+c

=b

i.e. average of a and c is equal to b.

Therefore, by the condition:

\frac{(n-2)+(5n+2)}{2}

2

(n−2)+(5n+2)

=4n - 1.

⇒\frac{n-2+5n+2}{2}

2

n−2+5n+2

=4n -1

⇒\frac{6n}{2}

2

6n

=4n-1

⇒3n=4n-1

⇒4n-3n=1

⇒n=1

Ans:The value of n is 1 and the numbers are 1-2 = -1 , 4*1-1=3 and 5*1+2 = 7 respectively as we see that -1+4=3 and 3+4=7.

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