Question

Insert 6 numbers in between 3 and 24 such that The resulting sequence is an AP

Answer 1

Given =>
a = 3
an= 24
n = 8
We know
an = a + (n-1) d
24 = 3 + (8-1) d
21 = 7d
d = 3

Therefore Ap becomes

3, 6,9,12,15,18,21,24....

Hope this helps you!

Answer 2

Given:

Arithmetic progression with first term 3 and last term 24.

To Find.

Insert 6 numbers between 3 and 24.

Solution:

Let 6 numbers between 3  and 24 be

3+d, 3+2d, 3+3d, 3+4d, 3+5d and 3+6d

As we insert 6 numbers between so 24 will be the 8th term of A.P.

T₈ = a + 7d = 24

3 + 7d = 24

7d = 24 - 3

d = 21/7

d = 3

put value of d = 3 we will get terms,

a + d = 3 +3 = 6

3+2d = 3 + 6 = 9

3+3d = 3 + 9 = 12

3+4d = 3 + 12 = 15

3+5d = 3 + 15 = 18 and

3+6d = 3 + 18 = 21

Hence, the six numbers between 3 and 24 will be 3, 6, 9, 12, 15, 18, 21.