Question

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Question :- The pth term of an A.P is 1/7( 2p - 1 ) . Find the sum of its first n terms.

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Answer 1

Given: ap= 1/7(2p-1)

Then, a1=1/7(2-1)=1/7

a2=1/7(4-1)=3/7

a3=1/7(6-1)=5/7

Therefore, d=a2-a1= 3/7-1/7= 2/7

Now, Sn= n/2(2a+(n-1)d)

Sn= n/2(2*1/7+(n-1)2/7)

=n/2(2/7+(n-1)2/7)

=n/2(2/7+2n/7-2/7)

=n/2(2n/7)-------- +2/7-2/7 gets cancelled

=2n^2/2*7

=n^2/7

Therefore, Sn=n^2/7
Given: ap= 1/7(2p-1)

Then, a1=1/7(2-1)=1/7

a2=1/7(4-1)=3/7

a3=1/7(6-1)=5/7

Therefore, d=a2-a1= 3/7-1/7= 2/7

Now, Sn= n/2(2a+(n-1)d)

Sn= n/2(2*1/7+(n-1)2/7)

=n/2(2/7+(n-1)2/7)

=n/2(2/7+2n/7-2/7)

=n/2(2n/7)-------- +2/7-2/7 gets cancelled

=2n^2/2*7

=n^2/7

Therefore, Sn=n^2/7

Answer 2

The pth term of the AP is 1/7 (2p-1). That is,
               ap=1/7 (2p-1)

Then the first term a1 will be,
               a1=1/7 [2(1) -1]
                   =1/7 (2-1)
                   =1/7
IIIly,        a2=1/7 [2(2) -1]
                   =3/7

∴ Common difference, d= a1-a2= 3/7-1/7
                                                         =2/7

Now we have; a1= 1/7 , d=2/7

                          an= a + (n-1) d
                              = 1/7 + (n-1) 2/7
                              = 1/7 + 2n/7 - 2/7
                              = 1/7 (1 + 2n -2)                      | Taking 1/7 as common 
                              = 1/7 (2n - 1)
                            
                         Sn= n/2 (a + an)
                              = n/2 [1/7 + 1/7 (2n-1)]
                              = n/2 (1/7 + 2n/7 - 1/7)
                              = n/2 (2n/7)
                              = 2n²/14
                              = n²/7