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Two A.P.’s are given 9, 7, 5, .... and 24, 21, 18, ... . If nth term of both the progressions are equal, then find the value of n and nth term.

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Sol Case (I)
        In an A. P.,
        9, 7, 5, .................
∴     a = 9 and d = 7 – 9 = –2
        tn = a + (n –1)d
        tn = 9 + (n –1) (–2)
        tn = 9 – 2n + 2
        tn = 11 – 2n                        ...(I)
        Case (II)
         In an A. P.,
         24, 21, 18, ................
         a = 24, d = 21 – 24 = –3
         tn = a + (n –1)d
         tn = 24 + (n –1) (–3)
         tn = 24 – 3n + 3
         tn = 27 – 3n                       ...(II)
        nth term of both the A. P. is same
        From (I) and (II)
        11 – 2n = 27 – 3n
      –2n + 3n = 27 – 11
                  n = 16                      ...(III)
                 tn = 11 – 2n [From (I)]
                     = 11 – 2 × 16 [From (III)]
                     = 11 – 32
                 tn = –21
∴     The value of n = 16 and the nth term is –21 which will be same for both the progressions.

Arithmetic Progression September 28 , 2018 0 Comments 11 views