Question

The sum of first 20 terms of an A.P. is one third of the sum of next 20 terms. If first term is 1 , then find the sum of the first 30 terms?

Plz guys ans... its urgent!!

Answer 1

Solution:

Given: The sum of first 20 terms of an AP is one-third of the sum o next 20 terms.
which means,   $S_{20}=(S_{40}-S_{20})/3$
The first term is 1 
which means,  $a=1$

To find: sum of first 30 terms ($S_{30}$)

By using the formula,$S_n=n(a+l)/2$,

          $S_{20}=(S_{40}-S_{20})/3$
          $20(1+a_{20})/2=(40(1+a_{40})/2-20(1+a_{20})/2)/3$
          [tex]10(1+a_{20})=[(20(1+a_{40})-10(1+a_{20})]/3 [/tex]                                                                                                                                                       [tex]10(1+a_{20})=10[2(1+a_{40})-(1+a_{20})]/3 [/tex]
         
          $(1+a_{20})=[2+2a_{40}1-a_{20}]/3$
          [tex]3(1+a_{20})=1+a_{40}-a_{20} [/tex]
          $1+a_{20}=1+a_{40}-a_{20}/3$
          $3+3a_{20}=1+a_{40}-a_{20}$
          [tex]2+4a_{20}=a_{40} [/tex]
          [tex]4a_{20}=a_{40}-2 [/tex]
          $a_{20}=(a_{40}-2)/4$ ....(i)

          $a+19d=a+39d-2$
          $1+19d=1+39d-2$
          [tex]1+19d=39d-1 [/tex]
          $1+1=39d-19d$
          $20d=2$
          $d=1/10$  .....(ii)

          $a_{30} =a +29d$
          $1+29/10=39/10=3.9$ =>$a_{30}=3.9$ ....(iii)

         $S_n=n(a+l)/2$
         $S_{30}=30(1+3.9)/2$
         [tex]S_{30}=15(4.9) [/tex]
         ∴$S_{30}=73.5$      Hope it helps