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Find the number of terms in a GP given that its first and last terms are 5 1/3 k and 243/256 k respectively and that its common ratio is 3/4.

Mathematics

ScanSolve AI Answer

Answer

7

Step-by-step solution

5 steps

The formula for the nn-th term (last term) of a Geometric Progression (GP) is L=arn−1L = ar^{n-1}, where aa is the first term, rr is the common ratio, and nn is the number of terms.

  1. Identify the given values and convert the first term to an improper fraction.

    First term, a=513k=5×3+13k=163ka = 5\frac{1}{3}k = \frac{5 \times 3 + 1}{3}k = \frac{16}{3}k.
    Last term, L=243256kL = \frac{243}{256}k.
    Common ratio, r=34r = \frac{3}{4}.

  2. Substitute these values into the formula L=arn−1L = ar^{n-1}.

    243256k=163k(34)n−1\frac{243}{256}k = \frac{16}{3}k \left(\frac{3}{4}\right)^{n-1}

  3. Simplify the equation by dividing both sides by kk (assuming k≠0k \neq 0) and isolating the term with nn.

    243256=163(34)n−1\frac{243}{256} = \frac{16}{3} \left(\frac{3}{4}\right)^{n-1}

    Multiply both sides by 316\frac{3}{16}:

    (34)n−1=243256×316\left(\frac{3}{4}\right)^{n-1} = \frac{243}{256} \times \frac{3}{16} (34)n−1=243×3256×16\left(\frac{3}{4}\right)^{n-1} = \frac{243 \times 3}{256 \times 16} (34)n−1=7294096\left(\frac{3}{4}\right)^{n-1} = \frac{729}{4096}

  4. Express the right side of the equation as a power of 34\frac{3}{4}.

    We know that 36=7293^6 = 729 and 46=40964^6 = 4096.
    So, 7294096=(34)6\frac{729}{4096} = \left(\frac{3}{4}\right)^6.
    Therefore, the equation becomes:

    (34)n−1=(34)6\left(\frac{3}{4}\right)^{n-1} = \left(\frac{3}{4}\right)^6

  5. Equate the exponents to solve for nn.

    n−1=6n-1 = 6 n=6+1n = 6 + 1 n=7n = 7

    The number of terms in the GP is 7.

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