Question

AP

Prove that,
S30 =3 (S20-S10)

Answer 1

here above is your answer mate

hope it helps you❤❤

Answer 2

Let sum of the first n terms of an A.P

Sn = ( n / 2) [ 2a + ( n -1)d ]

Now

S30 = ( 30 / 2) [ 2a + ( 30 -1)d ] let is eq ---------(1)

S20 - S10 = ( 20 / 2) [ 2a + ( 20 -1)d ] - ( 10 / 2) [ 2a + ( 10 -1)d ] S20 - S10 = 10 [ 2a + ( 20 -1)d ] - 5 [ 2a + ( 10 -1)d ] S20 - S10 = 10a + 190d - 45d .

3(S20 - S10 ) = 3(10a + 145d ) 3(S20 - S10 ) = 3 x 5(2a + 29d ) 3(S20 - S10 ) = ( 30 / 2) [ 2a + ( 30 -1)d ] let is eq-------(2)

From (1) and (2) we get

S30 = 3(S20 - S10 )

hope u got ur ans