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If sum of 3rd and 8th terms of an A.P. is 7 and sum of 7th and 14th terms is  –3, then find the 10th term.

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Ans.    For a given A. P.;
           tn = a + (n –1)d
           Where a = first term
                       d = common difference
           Sum of 3rd and 8th terms of an A.P. is 7
∴        t3 + t8 = 7
           a + (3 –1)d + (a + (8 –1)d  = 7
                            a + 2d + a + 7d = 7
                                        2a + 9d = 7                               ...(I)
           Sum of 7th and 14th terms is –3
∴                                   t7 + t14 = –3
           a + (7 – 1)d + a + (14 – 1)d = –3
                           a + 6d + a + 13d = –3
                                       2a + 19d = –3                              ...(II)
          Subtracting eqn. (I) From (II)

                2a + 9d     =  7
–              2a + 19d   = –3
          (–)    (–)          (+)


                        –10d =  10
                              d = –1
          Place d = –1 in eqn. (I)
           2a + 9(–1) = 7
                 2a – 9 = 7
                       2a = 7 + 9
                       2a = 16
                         a = 8
         t10 = a + (10 –1)d
         t10 = 8 + 9 × –1
         t10 = 8 – 9 = –1
∴      The 10th them is –1

Arithmetic Progression September 28 , 2018 0 Comments 74 views