Question

the sum of three numbers in A.P is 21 and the product of the first and the third terms exceeds the second term by 6 find the three numbers
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Answer 1

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$\huge\fcolorbox{red}{white}{Given:-}$

✴️Sum of first three terms of A.P = 21

✴️Product of first and third term - second term = 6

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$\huge\fcolorbox{red}{white}{Solution:-}$

Let the first term of A.P = a

Let the common difference be = d

Therefore,

• First term = a
• Second term = (a+d)
• Third term = ( a+2d)

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✴️Sum of n terms of A.P = \frac{n}{2}(2a + (n-1)d )

===> $\frac{3}{2}$ ( 2a + 2d ) = 21

===> $\frac{3}{2}$x 2(a+d) = 21

===> a + d = 7 ...................(i)

===> a = (7-d) ...................(ii)

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Now,

✴️(First term x third term) - second term = 6

===> [ a( a+2d) ] - ( a+d) = 6

===> [ a( a+2d) ] - 7 = 6 ; [ from (i) ]

===> [ a( a+2d) ] = 6+7

Putting the value of a from (ii) in this equation.

===> (7-d)(7-d+2d) = 13

===> (7-d)(7+d) = 13

===> 49 - d² = 13

===> d² = 49-13

===> d² = 36

===> d = 6

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Since,

The common difference = 6

First term will be (7-d) = (7-6) = 1

Hence,

• First term = 1
• Second term = (1+6) = 7
• Third term = ( 1+2x6) = 13

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