Question

If Sn denotes the sum of first n terms of an A.P., prove that S12=3(S8-S4).​

Answer 1

Answer:

Prove of  $S_1_2 = 3(S_8 - S_4)$  is shown below.

Step-by-step explanation:

We know that Sum of n terms of an A.P. , $S_n$ = $\frac{n}{2}[2a + (n-1)d]$ , where a = first term of AP and d = common difference.

To Prove: $S_1_2 = 3(S_8 - S_4)$

L.H.S =  $S_1_2$  =  $\frac{12}{2}[2a + (12-1)d]$

             $S_1_2$  = $6[2a + 11d]$

             $S_1_2$  =  $12a + 66d$

R.H.S =  $3(S_8 - S_4)$ = $3(\frac{8}{2}[2a + (8-1)d] - \frac{4}{2}[2a + (4-1)d])$

                              = $3(4[2a + 7d] - 2[2a + 3d])$

                              = $3(8a + 28d] - 4a - 6d])$

                              = $3( 4a + 22d)$ = $12a+66d$

Therefore, L.H.S = R.H.S hence proved that  $S_1_2 = 3(S_8 - S_4)$ .

Answer 2

Answer:

Here is ur attached answer

mark as brainliest