Question

the sum of three terms of an a.p is 21 and the product of first and third term exceeds the second term by 6 find three terms
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Answer 1

Answer :

1 , 7 , 13

Solution :

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