• Q. First term is of arithmatic progression sequence is 47 and the difference between consecutive numbers is 14 , Find the sum numbers from 36 to 40 terms.
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  • Term at 36 position is 47 + (36 - 1)x 14

    = [47 + (35)x 14 ]

    = [47 + 490]

    = [537]

    ie. S36 = (537)

  • Then,
    Sum of 36 to 40 position is (5/2)[2*537 + (5 - 1)x 14 ]

    = (2.5)[1074 + (4)x 14]

    = (2.5)[1074 + 56]

    = (2.5)[1130]

    ie. S36 to S40 = (2825)

  • Second method:-

    Term at 36 position is 47 + (36 - 1)x 14
    = 537 (=t1)

  • Term at 37 position is 47 + (37 - 1)x 14
    = 551 (=t2)

  • Term at 38 position is 47 + (38 - 1)x 14
    = 565 (=t3)

  • Term at 39 position is 47 + (39 - 1)x 14
    = 579 (=t4)

  • Term at 40 position is 47 + (40 - 1)x 14
    = 593 (=t5)

  • so the sequance of numbers from position 36 to 40 is 537, 551, 565, 579, 593

    = 537 + 551 + 565 + 579 + 593
    ie. S36 to S40 = 2825

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