:) Quick!How to solve this maths question?

The seventh term of an arithmetic progression is 12.The tenth term of the progression is greater than the second term by 8.Find the first term and the common difference,and hence,determine which term of the progression first exceeds 400.

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  • Anonymous
    9 years ago
    Favorite Answer

    Formula to use throughout the questtion:

    r th term of an A.P:

    A(r)=a+ (r-1)d

    Seventh term=12

    a+(7-1)d=12

    a+6d=12........(i)

    Tenth term = Second term + 8

    a+(10-1)d =a+(2-1)d + 8

    a + 9d = a +d + 8

    8d = 8

    d= 1

    Substitutte in (i)

    a + 6(1)=12

    a = 6

    To find r such that rth term exceeds 400

    A(r) > 400

    a + (r-1)d > 400

    6 + (r-1)1 > 400

    6 + r -1> 400

    r > 400 -5

    r > 395

    r = 396, 397,398.....and so on.....

    So the least integral value of r is 396.....

    YOUR ANSWER IS " 396 "

  • 9 years ago

    396

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