Maths

the first three terms of an ap are x, (3x +1) and (7x -4). Find the value x and 10th term

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  1. In A.P.

    a1 = initial term of arithmetic progression

    d = common difference

    nth term of A.P.

    an = a1 + ( n - 1 ) d

    In this case a1 = x , a2 = 3 x + 1 , a3 = 7 x - 4 so:

    a1 = x

    a2 = a1 + ( 2 - 1 ) d

    a2 = a1 + d

    3 x + 1 = x + d

    a3 = a1 + ( 3 - 1 ) d

    a3 = a1 + 2 d

    7 x - 4 = x + 2 d

    Now you must solve system:

    3 x + 1 = x + d

    7 x - 4 = x + 2 d

    Ttry it.

    The solutions are:

    x = 3 , d = 7

    a1 = x = 3

    an = a1 + ( n - 1 ) d

    a10 = a1 + ( 10 - 1 ) d

    a10 = a1 + 9 d

    a10 = 3 + 9 ∙ 7 = 3 + 63 = 66

    Your A.P.

    3 , 10 , 17 , 24 , 31 , 38 , 45 , 52 , 59 , 66 ...

    Proof:

    a1 = x = 3

    a2 = 3 x + 1

    10 = 3 ∙ 3 + 1 = 9 + 1 = 10

    a3 = 7 x - 4

    17 = 7 ∙ 3 - 4 = 21 - 4 = 17

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  2. please solve it clear

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  3. [(7 x - 4 = x + 2 d) - (3 x + 1 = x + d)] => 4 x - 5 = d

    substitute the value of d(4 x - 5) into any of the equations (7 x - 4 = x + 2 d) or (3 x + 1 = x + d)

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  4. Nice work

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