Question

36.
If Sn denotes the sum of the first n terms of an AP, prove that S30 = 3 (S20 - S10).
OR
The sum of the first 7 terms of an AP is 63 and the sum of its next 7 terms is 161. Find the 28th term
of this AP.
pls don't copy from Google ....​

Answer 1

Answer: 57

Step by step explanation:

Formula to be used:

Sn = ( n / 2) [ 2a + ( n - 1)d ]

Sum of the first 7 terms of an A.P is 63

( 7 / 2) [ 2a + 6d ] = 63

2a + 6d = 63 × 2/7

 2a + 6d = 18 ___(1)

Sum of its next 7 terms = 161. (Given)

Sum of first 14 terms = sum of first 7 terms + sum of next 7 terms.

S14 = 63 + 161 = 224

( 14 / 2) [ 2a + 13d ] = 224.

 7 [ 2a + 13d ] = 224.

[ 2a + 13d ] = 32 ___(2)

Simplifying (1) and (2), we get,

2a + 13d = 32

- 2a + 6d = 18

____________

7d = 14

d = 2 

.°. 2a + 13(2) = 32

2a + 26 = 32

a = 6/3

a = 3.

Now, To find 28th term,

t28 = a + ( 28 - 1) d

t28 = 3 + ( 28 - 1) 2

t28 = 57.

28th term = 57.

Thus, 28th term of A.P is 57.

Answer 2

To prove:
S30 = 3( S20 - S10 )
Where Sn = n/2( 2a + ( n - 1 )d )
So,
30/2(2a + (30-1)d ) = 3 [20/2(2a + (20-1)d ) - 10/2(2a + (10-1)d )]

15(2a + 29d) = 3 [10(2a + 19d) - 5(2a + 9d)]

*bring 3 to the other side*
*open the brackets in RHS*

5(2a + 29d) = 20a + 190d - 10a - 45d
10a + 145d = 10a + 145d

Hence proved