Login

Register Now

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Morbi adipiscing gravdio, sit amet suscipit risus ultrices eu. Fusce viverra neque at purus laoreet consequa. Vivamus vulputate posuere nisl quis consequat.

Register Now

Lost Password

Lost your password? Please enter your username and email address. You will receive a link to create a new password via email.

The A.P. in which 4th term is –15 and 9th term is –30. Find the sum of the first 10 numbers.

Print or Save

Sol.      Let ‘a’ be the first term and ‘d’ the common difference.

           Given : t4 = –15
           For t4 put n = 4
∴        a + (4 – 1)d = –15
∴                a + 3d = –15                  ...(I)
                         t9 = a + (9 – 1)d
          For t9 put n = 9
∴               a + 8d = –30                   ...(ii)
         Subtracting eq. (I) from (II),

         a   +   8d   =   –30
  –     a   +   3d   =   –15
     (–)   (–)           (+)


                   5d   =   –15
∴                  d   =   –3
        Substituting d = –3, in eqn (I),
         a + 3 × (–3)  =  15
                 a – 9    = –15
                          a = –15 + 9
∴                        a = –6
∴     a = –6 and d = –3
       Sum of first 10 terms,

      Sn\({n\over2}[2a+(n-1)d]\)

       For S10 put n = 10
       S10 =  \(10\over 2\) [ 2(–6) + (10 – 1)(–3)]
       S10 =  5[–12 + 9 (–3)]
       S10 =  5 [–12 –27]
       S10 = –195
∴    Sum of first ten numbers =  –195

∴    Sum of first ten numbers = –195

Arithmetic Progression September 28 , 2018 0 Comments 40 views