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In the following APs find the missing term in the boxes. Q.3 (all)

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The best way to solve all the problems of arithmetic progressions of exercise 5.2 of class 10th math, How i solve easily to this question In the following APs find the missing term in the boxes.

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  1. (i) For the given A.P., 2,2 , 26

    The first and third term are;

    a = 2

    a3 = 26

    As we know, for an A.P.,

    an = a+(n −1)d

    Therefore, putting the values here,

    a3 = 2+(3-1)d

    26 = 2+2d

    24 = 2d

    d = 12

    a2 = 2+(2-1)12

    = 14

    Therefore, 14 is the missing term.

    (ii) For the given A.P., , 13, ,3

    a2 = 13 and

    a4 = 3

    As we know, for an A.P.,

    an = a+(n−1) d

    Therefore, putting the values here,

    a2 = a +(2-1)d

    13 = a+d ………………. (i)

    a4 = a+(4-1)d

    3 = a+3d ………….. (ii)

    On subtracting equation (i) from (ii), we get,

    – 10 = 2d

    d = – 5

    From equation (i), putting the value of d,we get

    13 = a+(-5)

    a = 18

    a3 = 18+(3-1)(-5)

    = 18+2(-5) = 18-10 = 8

    Therefore, the missing terms are 18 and 8, respectively.

    (iii) For the given A.P.,

    = 5 and

    a4 = 19/2

    As we know, for an A.P.,

    an = a+(n−1)d

    Therefore, putting the values here,

    a4 = a+(4-1)d

    19/2 = 5+3d

    (19/2) – 5 = 3d

    3d = 9/2

    d = 3/2

    a2 = a+(2-1)d

    a2 = 5+3/2

    a2 = 13/2

    a3 = a+(3-1)d

    a3 = 5+2×3/2

    a3 = 8

    Therefore, the missing terms are 13/2 and 8, respectively.

    (iv) For the given A.P.,

    a = −4 and

    a6 = 6

    As we know, for an A.P.,

    an = a +(n−1) d

    Therefore, putting the values here,

    a6 = a+(6−1)d

    6 = − 4+5d

    10 = 5d

    d = 2

    a2 = a+d = − 4+2 = −2

    a3 = a+2d = − 4+2(2) = 0

    a4 = a+3d = − 4+ 3(2) = 2

    a5 = a+4d = − 4+4(2) = 4

    Therefore, the missing terms are −2, 0, 2, and 4 respectively.

    (v) For the given A.P.,

    a2 = 38

    a6 = −22

    As we know, for an A.P.,

    an = a+(n −1)d

    Therefore, putting the values here,

    a2 = a+(2−1)d

    38 = a+d ……………………. (i)

    a6 = a+(6−1)d

    −22 = a+5d …………………. (ii)

    On subtracting equation (i) from (ii), we get

    − 22 − 38 = 4d

    −60 = 4d

    d = −15

    a = a2 − d = 38 − (−15) = 53

    a3 = + 2= 53 + 2 (−15) = 23

    a4 = a + 3d = 53 + 3 (−15) = 8

    a5 = a + 4d = 53 + 4 (−15) = −7

    Therefore, the missing terms are 53, 23, 8, and −7, respectively.

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