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Q1a. The 5th and 10th terms of an A.P. are 12 and -27 respectively. Find the A.P. and its 15th term. Q1b. If the first and last terms of an A.P., with d = 1 1/2 are 2 1/2 and 19 respectively, how many terms has the A.P.?

Mathematics

ScanSolve AI Answer

Answer

-42

Step-by-step solution

5 steps

  1. Set up equations for the given terms of the A.P.

    The formula for the nn-th term of an A.P. is Tn=a+(n−1)dT_n = a + (n-1)d, where aa is the first term and dd is the common difference.
    Given T5=−12T_5 = -12:

    a+(5−1)d=−12a + (5-1)d = -12 a+4d=−12(1)a + 4d = -12 \quad (1)

    Given T10=−27T_{10} = -27:

    a+(10−1)d=−27a + (10-1)d = -27 a+9d=−27(2)a + 9d = -27 \quad (2)

  2. Solve the system of equations to find the common difference dd.

    Subtract equation (1) from equation (2):

    (a+9d)−(a+4d)=−27−(−12)(a + 9d) - (a + 4d) = -27 - (-12) 5d=−27+125d = -27 + 12 5d=−155d = -15 d=−155d = \frac{-15}{5} d=−3d = -3

  3. Substitute the value of dd into equation (1) to find the first term aa.

    a+4(−3)=−12a + 4(-3) = -12 a−12=−12a - 12 = -12 a=−12+12a = -12 + 12 a=0a = 0

  4. State the A.P.

    The first term is a=0a=0 and the common difference is d=−3d=-3.
    The A.P. is 0,(0−3),(0−2×3),…0, (0-3), (0-2 \times 3), \dots
    The A.P. is 0,−3,−6,…0, -3, -6, \dots

  5. Calculate the 15th term (T15T_{15}).

    Using the formula Tn=a+(n−1)dT_n = a + (n-1)d:

    T15=0+(15−1)(−3)T_{15} = 0 + (15-1)(-3) T15=0+(14)(−3)T_{15} = 0 + (14)(-3) T15=−42T_{15} = -42

    The A.P. is 0,−3,−6,…0, -3, -6, \dots and its 15th term is −42\boxed{-42}.

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