Question

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

Answer 1

Answer:

The three terms are 1, 7 ,13 and 13, 7 , 13

Step-by-step explanation:

Given :  

Sum of three terms of an A.P. is 21

Let the three terms in AP are (a - d), a, (a + d)

Then, we have

(a – d) + a + (a + d) = 21

3a = 21

a = 21/3

a = 7 ………..(1)

Also it is given that the,product of the first and the third terms exceeds the second term by 6, therefore we have,  

(a – d) (a + d) = a + 6

a² – d² = a + 6

7² – d² = 7 + 6

[From eq 1, a = 7]

49 – d² = 13

d² = 36

d = √36

d = ±6

If d = 6 , then  

First term , ( a – d) = 7 – 6 = 1

Third term , (a + d) = 7 + 6 = 13

Second term ,a = 7  

Therefore, the three terms are 1, 7 and 13.  

 

If d = - 6 , then  

First term , ( a – d) = 7 – (-6) = 7 + 7 = 13

Third term , (a + d) = 7 + (-6) = 7 - 6 = 1

Second term ,a = 7  

Therefore, the three terms are 13, 7 and 13

Hence, the three terms are 1, 7 ,13 and 13, 7 , 13

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Answer 2

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