Question

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

Answer 1

Answer:

Follow the steps given below

Explanation:

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 this helps you

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Answer 2

Explanation:

Given : -

• The sum of an AP is 21 and product of th first .

• third term exceed the second term by 6.

To Find : -

• Find the 3 terms

Solution : -

• 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 

Substitute all values :

a² + 2ad = a + d + 6

a²+ 2a(7 - a) = a + (7 - a) + 6 

a² + 14a - 2a² = 13

a² - 14a + 13 = 0 

(a - 1)(a - 13) = 0

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