Math, asked by gagangowda85499, 1 year ago

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​​

Answers

Answered by TheCommando
69

Let the three terms of AP be

 {a}_{1} =  a \\ {a}_{2} = a + d \\ {a}_{3}  = a + 2d

Given:

 {a}_{1} + {a}_{2} + {a}_{3} = 21 \\ \implies a + a + d + a + 2d = 21 \\ \implies 3a + 3d = 21 \\ \implies a + d = 7 \\ \implies d = 7 - a \bold{(Equation 1)}

Also,

 {a}_{1} \times {a}_{3} = {a}_{2} + 6 \\ \implies (a) × (a + 2d) = (a + d) + 6 \\ \implies a^{2} + 2ad = a + d + 6

Putting value of d from Equation 1

 a^{2} + 2a(7-a) = a + (7-a) + 6 \\  \implies a^{2} + 14a - 2a^{2} = 13 \\ \implies -a^{2} + 14a = 13 \\ \implies a^{2} - 14a + 13 = 0 \\ \implies a^{2} - a - 13a + 13 = 0 \\ \implies a(a - 1) - 13(a - 1) = 0 \\ \implies (a-13)(a-1) = 0

a - 13 = 0

a = 13

a - 1 = 0

a = 1

Putting value of a in Equation 1

d = 7 - a

d = 7 - 1 = 6 or d = 7 - 13 = -6

If a = 1 and d = 6

Then,

 {a}_{1} =  a = 1 \\ {a}_{2} = a + d = 1 + 6 = 7 \\ {a}_{3}  = a + 2d = 1 + 2(6) = 13

If a = 13 and d = -6

 {a}_{1} =  a = 13 \\ {a}_{2} = a + d = 13 + (-6) = 7 \\ {a}_{3}  = a + 2d = 13 + 2(-6) = 1

Therefore, the terms are either 1, 7, 13..... or 13, 7, 1.....


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Answered by Anonymous
39

Solution

Let the AP in Form of

a - d , a and a + d

The sum of AP

a + d + a + a - d = 21

3a = 21

a = 7

The Product of 1st and 3rd term of the Ap is

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

( - ) = a + 6 ...... Equation

a = 7

PUTTING IT IN EQUATION

49 - d² = 7 + 6

(- d²) = - 49 + 13

d² = 36

d = ± 6

So the Ap Could be ...

We have a = 7 and d = 6

= (a-6), (a) ,(a + 6)

(1 , 7, 13 ,19...)

Now taking d = -6

The AP become ...

( 13 , 7 , 1 , -5 ...)

↪↪↪↪↪↪↪↪↪↪↪↪↪↪


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