Question

if the number n-2 , 4n-1 and 5n+2 are in AP , find the value of n.​

Answer 1

G I V E N :-

• The numbers n-2, 4n-1 and 5n+2 are in AP.

T O F I N D :-

• We have to find the value of n.

S O L U T I O N :-

• n-2, 4n-1 and 5n+2 are in AP.

We know that common difference will eb same across the numbers in a sequence. So,

☞ (4n-1) - (n-2) = (5n+2) - (4n-1)

☞ 4n - 1 - n + 2 = 5n + 2 - 4n + 1

☞ 4n + 4n - 5n - n - 1 + 2 - 2 - 1

☞ 8n - 6n = 0

☞ 2n - 2 = 0

☞ 2n = 2

☞ n = 1

Also, the numbers are as follows :

• 4n-1

☞ 4(1) - 1

☞4 - 1

☞ 3

• n-2

☞ (1) -2

☞1 - 2

☞ -1

• 5n+2

☞ 5(1) + 2

☞ 5 + 2

☞ 7

Answer 2

Given:

The numbers n-2, 4G I V E N :-

The numbers n-2, 4n-1 and 5n+2 are in AP.

To find:

We have to find the value of n.

Solution:

n-2, 4n-1 and 5n+2 are in AP.

We know that common difference will eb same across the numbers in a sequence. So,

☞ (4n-1) - (n-2) = (5n+2) - (4n-1)

☞ 4n - 1 - n + 2 = 5n + 2 - 4n + 1

☞ 4n + 4n - 5n - n - 1 + 2 - 2 - 1

☞ 8n - 6n = 0

☞ 2n - 2 = 0

☞ 2n = 2

☞ n = 1

Also, the numbers are as follows :

4n-1

☞ 4(1) - 1

☞4 - 1

☞ 3

n-2

☞ (1) -2

☞1 - 2

☞ -1

5n+2

☞ 5(1) + 2

☞ 5 + 2

☞ 7