Question

If $S_{n}$ denotes the sum of the first n terms of an A.P., prove that $S_{30}$ = 3($S_{20}$ - $S_{10}$)

Answer 1

Answer with Step-by-step explanation:

Given :  

Sn denotes the sum of the first n terms of an A.P

We have to prove that : S30 = 3(S20 - S10)

By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]

L.H.S :  

S30 = (30 / 2) [ 2a + ( 30 -1)d ]  

S30 = 15[2a + 29d]

S30 = 30a + 435d …………(1)

R.H.S :

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 = 10 [ 2a + 19d ] - 5 [ 2a + 9d ]

S20 - S10 = 20a + 190d  - 10a - 45d

S20 - S10 = 10a + 190d  -  45d  

S20 - S10 = 10a + 145d  

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

3(S20 - S10 ) = 3 x 5(2a + 29d )

3(S20 - S10 ) = 15(2a + 29d )

3(S20 - S10 ) = 30a + 435d ………(2)

From eq 1 & 2 ,  

L.H.S = R.H.S  

S30 = 3(S20 - S10 )

Hence Proved….

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Answer 2

Answer

L.H.S.

$\tt{S_{30} = \frac{30}{2}(2a + 19d)}$

= 15[2a + 19d]

= 30a + 435d

R.H.S.

$\tt{3(S_{20} - S_{10}) = 3(\frac{20}{2}(2a + 19d) - \frac{10}{2}(2a + 9d)}$

= 3[20a + 190d - 10a - 45d]

= 30a + 435d

L.H.S. = R.H.S.

Hence Proved!