Question

insert four numbers between 8 and 26, so that the resulting sequence is an A.P.

Answer 1

As we have to find an AP of 6 terms having first and last term as 8 and 26.
Firstly, we will find common difference.
So, We know An = A+(n-1)d 
Here, 26= 8+(6-1)d = 8+5d 
5d = 26-8 = 18
d=18/5.
So, The sequence will be 
8,8+18/5 ,8+2*18/5,8+3*18/5,8+4*18/5 ,26
i.e.8,58/5, 76/5,94/5,112/5,26 .

Answer 2

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Assume that A1, A2, A3, A4, and A5 are the five numbers between 8 and 26, such that the sequence of an A.P becomes 8, A1, A2, A3, A4, A5, 26

Here, a= 8, l =26, n= 5

Therefore, 26= 8+(7-1)d

Hence it becomes,

$\tt 26 = 8+6d$

$\tt 6d = 26-8 = 18$

$\tt 6d= 18$

$\tt d = 3$

$\tt A1= a+d = 8+ 3 =11$

$\tt A2= a+2d = 8+ 2(3) =8+6 = 14$

$\tt A3= a+3d = 8+ 3(3) =8+9 = 17$

$\tt A4= a+4d = 8+ 4(3) =8+12 = 20$

$\tt A5= a+5d = 8+ 5(3) =8+15 = 23$

Hence, the required five numbers between the number 8 and 26 are 11, 14, 17, 20, 23

Hope it's Helpful....:)