The sum of three terms of an AP is 12 The product of first and third term is 8 greater than second term find the terms
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The sum of three terms of an AP is 12 The product of first and third term is 8 greater than second term find the terms
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2 thoughts on “The sum of three terms of an AP is 12 The product of first and third term is 8 greater than second term find the terms<br />who wi”

  1. Given that, the sum of three terms of an AP is 12. & The product of first and third term is 8 greater than second term.

    So, Let’s consider a – d , a , a + d be three terms of A.P.

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    [tex]{\underline{\pmb{\frak{\bigstar\:According\:to\:the\:Question\::}}}}\\\\[/tex]

    • Sum of first three terms of AP is 12.

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    [tex]\dashrightarrow\sf (a – d) + (a) + (a + d) = 12\\\\\\ \dashrightarrow\sf 3a\: \cancel{- \: d} \: \cancel{+\: d} = 12\\\\\\ \dashrightarrow\sf 3a = 12\\\\\\ \dashrightarrow\sf a = \cancel{\dfrac{12}{3}}\\\\\\ \dashrightarrow{\underline{\boxed{\pmb{\frak{a = 4}}}}}\:\bigstar\\\\[/tex]

    [tex]\therefore\:{\underline{\sf{Second\:term\:(a)\:of\:AP\:is\:{\pmb{4}}.}}}[/tex]

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    ▪︎Now, Let’s solve second Condition

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    • The product of first and third term is 8 greater than second term.

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    [tex]\qquad\dashrightarrow\sf (a – d)(a + d) = a + 8\\\\\\ \qquad\dashrightarrow\sf (4 – d)(4 + d) = 4 + 8\\\\\\ \qquad\dashrightarrow\sf 4^2 – d^2 = 12\\\\\\ \qquad\dashrightarrow\sf 16 – d^2 = 12\\\\\\ \qquad\dashrightarrow\sf d^2 = 16 – 12\\\\\\ \qquad\dashrightarrow\sf d^2 = 4\\\\\\ \qquad\dashrightarrow{\underline{\boxed{\pmb{\frak{d = 2}}}}}\:\bigstar\\\\[/tex]

    [tex]\therefore\:{\underline{\sf{Common\:diffrence\:(d)\:between\:the\:terms\:of\:AP\:is\:{\pmb{2}}.}}}[/tex]

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    Therefore, The required terms of A.P. are :

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    [tex]\qquad\quad\twoheadrightarrow\sf (a – d) \:,\: a \:,\: (a + d)\\\\\\ \qquad\quad\twoheadrightarrow\sf (4 – 2) \:,\: 4 \:,\: (4 + 2)\\\\\\ \qquad\quad\twoheadrightarrow{\pmb{\frak{\purple{2 \:,\: 4 \:,\: 6}}}}[/tex]

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  2. Given :-

    The sum of three terms of an AP is 12 The product of first and third term is 8 greater than second term

    To Find :-

    Terms

    Solution :-

    Let the terms of the AP be a – d, a, a + d

    a – d + a + a + d = 12

    (a + a + a) + (d – d) = 12

    3a = 12

    12/3 = a

    4 = a

    Now

    (a – d) × (a + d) = a + 8

    (4 – d) × (4 + d) = 8 + 4

    (4 × 4) – (d × d) = 8 + 4

    16 – d² = 8 + 4

    -d² = 8 – 12

    -d² = -4

    d² = 4

    d = √4

    d = ±2

    Finding the term

    a – d = 4 – 2 = 2

    a = 4

    a + d = 4 + 2 = 6

    [tex]\\[/tex]

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