Question

If Sn demotes the sum of first n terms of an AP. Prove that S12=3(S8-S4)​

Answer 1

Answer:-

We know that sum of n terms of an AP sn = (n/2)

[2a + (n - 1) * d]

(i) S12 = (12/2)[2a + (12 - 1) * d]

= 6[2a + 11d]

= 12a + 66d

(ii) S8 = 8/2[2a + (8 - 1) * d]

= 4[2a + 7d]

= 8a + 28d

(iii) S4 = (4/2)[2a + (4 - 1) * d]

= 2[2a + 3d]

= 4a + 6d.

From (ii) & (iii)

⇒ S8 - S4 = 8a + 28d - 4a - 6d

= 4a + 22d.

Now,

⇒ 3(S8 - S4) = 3(4a + 22d)

= 12a + 66d

= S12.

RHS = LHS (Proved)

$\begin{gathered} \\ \end{gathered}$

________________________

Answer 2

$Here \: is \: your \: answer \: my \: dear$

Let 1

st

term of A.P=a

and common difference =d

We have to prove that

S

12

=3(S

−S

4

)

S

12

=

2

12

[2a+(12−1)d]

S

8

=

2

8

[2a+(8−1)d]

S

4

=

2

4

[2a+(4−1)d]

Now, R.H.S=3(S

8

−S

4

)

=3[

2

8

(2a+(8−1)d−

2

4

(2a+(4−1)d]

=3[8x+28d−4a−6d]

=6(2a+11d)

=

2

12

[2a+(12−1)d]

=S

12

=L.H.S