Question

1. The sum of three terms of an A.P. is 21 and the product of the first and the third term
exceeds the second term by 6, find three terms.​

Answer 1

Answer:

1 , 7 , 13

Step-by-step explanation:

Let the three terms of AP be ;

(a - d) , a , (a + d) .

Now

According to the question ,

The sum of the three terms of the AP is 21.

Thus ,

=> (a - d) + a + (a + d) = 21

=> 3a = 21

=> a = 21/3

=> a = 7

Also ,

It is given that ,

The product of 1st and 3rd term exceeds the 2nd term by 6 .

Thus ,

=> (a - d)(a + d) = a + 6

=> a² - d² = a + 6

=> d² = a² - a - 6

=> d² = 7² - 7 - 6

=> d² = 49 - 13

=> d² = 36

=> d = √36

=> d = 6

Now ,

1st term = (a - d) = 7 - 6 = 1

2nd term = a = 7

3rd term = a + d = 7 + 6 = 13

Hence ,

Required terms are : 1 , 7 , 13

Hope this help you