Question

plz ans it...as soon as possible..

Answer 1

We know that Tn = a + (p - 1) * d

                             = 3p - 1/6

Now,

T1 = 3(1) - 1/6

     = 2/6

     = 1/3


T2 = 3(2) - 1/(6)

      = 5/6



First term T1 = 1/3.

Common Difference d = T3 - T2

                                       = 1/2.



We know that number of terms Tn = a + (n - 1) * d

                                                           = 1/3 + (n - 1) * 1/2

                                                           = (3n - 1)/6.


Therefore Sn = n/2(T1 + Tn)

                       = n/2(1/3 + (3n - 1)/6)

                       = n/2 * (3n + 1)/6

                       = (3n + 1)/12.


Therefore the answer is Option - (B).



Hope this helps!

Answer 2

Here is your answer ,

Given,

pth term = ( 3p - 1 ) / 6

Let , p = 1

=> 1st term = ( 3 × 1 - 1 ) / 6

=> 1st term = ( 3 - 1 ) / 6

=> 1st term = 2/6

•°• 1st term = ( 1/3 )

Let, p = 2

=> 2nd term = ( 3 × 2 - 1 ) / 6

=> 2nd term = ( 6 - 1 ) / 6

•°• 2nd term = 5/6

Now,

=> Common difference = 2nd term - 1st term

= ( 5/6 ) - ( 1/3 )

= ( 5 - 2 ) / 6

= 3/6

= ( 1/2 )

First term ( a ) = ( 1/3 )

Common difference ( d ) = ( 1/2 )


Now,

=> Sum of first n terms = ( n/2 ) { 2a + ( n - 1 )d }

= ( n/2 ) [ 2( 1/3 ) + ( n - 1 ) ( 1/2 ) ]

= ( n/2 ) [ ( 2/3 ) + ( n/2 ) - ( 1/2 )

= ( n/2 ) [ { ( 2/3 ) - ( 1/2 ) } + ( n/2 ) ]

= ( n/2 ) [ { ( 4 - 3 ) / 6 } + ( n/2 ) ]

= ( n/2 ) [ ( 1/6 ) + ( n/2 ) ]

= ( n/2 ) [ ( 1 + 3n ) / 6 ]

= ( n / 12 ) ( 3n + 1 )

b. ) ( n / 12 ) ( 3n + 1 ).