Question

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

Answer 1

let x be the first term and a be ap coefficient
⇒x,x+a,x+2a,x+3a,.....,x+(n-1)a
Sum of terms of a AP=n(1st term+last term)/2
so
S12=12(x+x+(12-1)a)/2
       =6(2x+11a)
S8= 8(x+x+(8-1)a)/2
    = 4(2x+7a)
S4=4(x+x+(4-1)a)/2
    =2(2x+3a)
S8-S4= 4(2x+7a)-2(2x+3a)
          = 8x+28a-4x-6a
          = 4x+22a
3(S8-S4)=3(4x+22a)
               =3(2(2x+11a))
               =6(2x+11a)=S8
⇒3(S8-S4)=S8

Answer 2

Solution :

Let the first term be a and common difference be d.

We have to prove $\sf S_{12} = 3(S_{8} - S_{4})$

In RHS, we have : $\sf 3(S_{8} - S_{4})$

Now,

$\underline{\boxed{\sf S_{n} = \dfrac{n}{2} . [2a+(n-1)d]}}$

$\sf : \implies 3 \Bigg[\dfrac{8}{2}(2a+(8-1)d) - \dfrac{4}{2}(2a+(4-1)d)\Bigg]$

$\sf : \implies 3 \Bigg[\cancel{\dfrac{8}{2}}(2a+7d) - \cancel{\dfrac{4}{2}}(2a+3d)\Bigg]$

$\sf : \implies 3 [4(2a+7d) - 2(2a+3d)]$

$\sf : \implies 3 \times 2[2(2a+7d) - (2a+3d)]$

$\sf : \implies 6(4a+14d - 2a+3d)$

$\sf : \implies 6(2a+11d)$

In LHS, we have : $\sf S_{12}$

Now,

$\underline{\boxed{\sf S_{n} = \dfrac{n}{2} . [2a+(n-1)d]}}$

$\sf : \implies \dfrac{12}{2} (2a + (12-1)d)$

$\sf : \implies \cancel{\dfrac{12}{2}} (2a + 11d)$

$\sf : \implies 6 (2a + 11d)$

As, LHS = RHS,

Hence, Proved.