• Q. First term is of geometric progression sequence is 3 and the common ratio numbers is 3 , Find the sum numbers from 5 to 9 terms, and sum of nth term.
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  • solve:-
    Term at 5 position is 3 *3(5-1)
    = 3 *3(4)
    = 3 * 81
    = 243


  • Now,
    The sum of 5 to 9 position is 243*[1 - 3(5)] /[1-3]
    = (243)[ 1 - 3(5)]/[1-3]

    = (243)(-242)/(-2)

    = (243)(121)

    ie. S5 to S9 = 29403


  • Second method:-
    Term at 5 position is 3 *3(5-1)
    = 3 *3(4)

    = 3 * 81
    = 243 (=t1)


  • Term at 6 position is 3 *3(6-1)
    = 3 *3(5)

    = 3 * 243
    = 729 (=t2)


  • Term at 7 position is 3 *3(7-1)
    = 3 *3(6)

    = 3 * 729
    = 2187 (=t3)


  • Term at 8 position is 3 *3(8-1)
    = 3 *3(7)

    = 3 * 2187
    = 6561 (=t4)


  • Term at 9 position is 3 *3(9-1)
    = 3 *3(8)

    = 3 * 6561
    = 19683 (=t5)


  • so the sequance of numbers from position 5 to 9 are 243, 729, 2187, 6561, 19683
    = 243 + 729 + 2187 + 6561 + 19683;

    = 29403

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