Question

sum of three terms of an ap is 21 and the product of first and the third term exceeds the second term by 6 find three terms

Answer 1

a – d) + a + (a + d) = 21 = 3a, so a was 7
a^2 – d^2 = a + 6
d^2 = 49 – 13 = 36, so d was 6
Terms: 1, 7, 13

Answer 2

Answer:

Step-by-step explanations:

Let the three terms will be:

a-d,a,a+d

And it is given that , their sum is 21

Now, a-d+a+a+d=21

So, we get a=7..

Given: (a-d)•(a+d)= a+6

a^2 - d^2= 13

d=6

Now the A.P. will be 1,7,13