Question

In A.P, prove that S30 = 3(S20 - S10 )

Answer 1

sum of the first n terms of an A.P= ( n / 2) [ 2a + ( n -1) d ]

so, S30=( 30 / 2) [ 2a + ( 30 -1) d ]
           = 15 [ 2a + 29d ]________________________________________(1)

now, S20 - S10 = ( 20 / 2) [ 2a + ( 20 -1)d ] - ( 10 / 2) [ 2a + ( 10 -1) d ]
                       = 10 [ 2a + ( 20 -1)d ] - 5 [ 2a + ( 10 -1)d ]
                       = 10a + 190d  -  45d
                       = 10a + 145d
therefore, 3( S20 - S10 ) = 3( 10a + 145d )
                                    = 3 x 5 ( 2a + 29d )
                                    =15 (2a + 29d)____________________________(2)

Hence,from 1 and 2,we proved that S30 = 3(S20 - S10 )

Answer 2

Answer:

sum of the first n terms of an A.P= ( n / 2) [ 2a + ( n -1) d ]

so, S30=( 30 / 2) [ 2a + ( 30 -1) d ]

         = 15 [ 2a+29d] _____________________(1)

now, S20 - S10 = ( 20 / 2) [ 2a + ( 20 -1)d ] - ( 10 / 2) [ 2a + ( 10 -1) d ]

                       = 10 [ 2a + ( 20 -1)d ] - 5 [ 2a + ( 10 -1)d ]

                       = 10a + 190d  -  45d

                       = 10a + 145d

therefore, 3( S20 - S10 ) = 3( 10a + 145d )

= 3 x 5 ( 2a +29d )             =15(2a+29d)_______________(2)

Hence,from 1 and 2,we can see that S30 = 3(S20 - S10)

Plz mark BRAINLIEST

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