Question

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

Answer 1

Step-by-step explanation:

Let a A.P is defined such that the the first term is a and the common difference is d

The sum n terms of am A.P is given by

$S_n=\dfrac{n}{2}(2a+(n-1)d)\\\\S_{12}=\dfrac{12}{2}(2a+(12-1)d)\\\\S_{12}=12a+66d$

Similarly

$S_n=\dfrac{n}{2}(2a+(n-1)d)\\\\S_{8}=\dfrac{8}{2}(2a+(8-1)d)\\\\S_{8}=8a+28d$

$S_n=\dfrac{n}{2}(2a+(n-1)d)\\\\S_{4}=\dfrac{4}{2}(2a+(4-1)d)\\S_4=4a+6d$

Now we have

$3(S_8-S_4)=3(8a+28d-4a-6d)=12a+66d$

So From above It is proved that

$s_{12}=3(S_8-S_4)$

Answer 2

$\Large \bf Solution :$

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

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

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

Now,

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

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

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

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

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

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

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

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

Now,

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

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

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

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

$\bf As, \: LHS = RHS,$

$\underline{\underline{\bf Hence, \: Proved.}}$