Question

The sum of first three terms of an AP is 21 and the product of the first and the third term exceeds the second term by 6. Find the

Answer 1

Answers

Let a = 1st term, d = common difference 

Then the 1st 3 terms are a, a + d, a + 2d 

Their sum is 3a + 3d = 21, so a + d = 7, and 

thus d = 7 - a. 

The problem says a(a + 2d) = a + d + 6 

a^2 + 2ad = a + d + 6 Substitute: 

a^2 + 2a(7 - a) = a + (7 - a) + 6 

a^2 + 14a - 2a^2 = 13, or 

a^2 - 14a + 13 = 0 

(a - 1)(a - 13) = 0, so a = 1 or a = 13 

If a = 1, d = 6; if a = 13, d = - 6. So there 

are 2 possible sequences: 1, 7, 13... or 13, 7, 1

hope it helps

Answer 2

hey here is your answer