Question

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

Answer 1

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...

Answer 2

Heya mate! ✌

Here is your answer

Let the three terms of AP are a,a+d,a+2d.

Now,

As per the question ,

a+a+d+a+2d = 3a+3d = 21

i.e. a+d = 7 or, d= 7-a.

Now,

a(a+2d)-(a+d) = 6

a{a+2(7-a)}-7 = 6

a{a+14-2a}=13

a{14-a} =13

14a-a²=13

a²-14a +13 = 0

By applying quadratic formula, you will get

a= 13, 1.

d will be 7-a i.e. -6 ,6.

Now, there will be two AP

13,7, 1 and 1,7,13.

Answer....

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