Question

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

Answer 1

TO PROVE :–

$\\ \implies \bf S_{12} = 3(S_{8} - S_{4})$

SOLUTION :–

• We know that Sum of n terms of A.P. –

$\\ \implies { \boxed{ \bf S_{n} = \dfrac{n}{2} \left[2a + (n - 1)d \right]}}$

• Let's take L.H.S. –

$\\ \implies \bf S_{12} = \dfrac{12}{2} \left[2a + (12 - 1)d \right]$

$\\ \implies \bf S_{12} = 6 (2a + 11d )$

$\\ \implies \bf S_{12} = 12a + 66d \: \: \: \: - - - eq.(1)$

• Let's take R.H.S. –

$\\ \implies \bf 3(S_{8} -S_{4}) = 3 \left( \dfrac{8}{2} \left[2a + (8- 1)d \right] - \dfrac{4}{2} \left[2a + (4 - 1)d \right] \right)$

$\\ \implies \bf 3(S_{8} -S_{4} )=3 \left( 4 \left[2a + (8- 1)d \right] - 2 \left[2a + (4 - 1)d \right] \right)$

$\\ \implies \bf 3(S_{8} -S_{4}) = 12(2a + 7d )- 6 (2a + 3d)$

$\\ \implies \bf 3(S_{8} -S_{4}) =24a + 84d - 12a - 18d$

$\\ \implies \bf 3(S_{8} -S_{4}) =12a + 66d \: \: \: \: - - - eq.(2)$

▪︎ By eq.(1) & eq.(2) –

$\\ \implies \bf L.H.S. =R.H.S.$

$\\ \dashrightarrow { \underline{ \: \bf Hence \: \: proved }}$

Answer 2

★Answer:-

$\bf \ \: S_n \: is \: sum \: of \: first \: n \: terms \: of \: an \: A.P .$

$\bf \ \: let \: a \: be \: \: the \: term \: and \: d \: be \: the \: common \: difference \: of \: the \: given \: A.P$

Then,

$\sf \ \: S_n = \frac{n}{2} [2a + (n - 1)d]$

$\sf \ \: S_1_2= \frac{12}{2} [ 2a + (12 - 1)d]$

$S_1_2=6[2a + 11d ]$

$S_1_2=12a + 66d \: ........(1)$

Now,

$S_8 = \frac{8}{2} [2a + (8 - 1)d] \\ S_8 = 4[2a + 7d] \\ S_8 = 8a + 28d...........(2)$

and,

$S_4 = \frac{8}{2} [2a + (4 - 1)d] \\ S_4 = 2[2a + 3d] \\ S_4 = 4a + 6d \: .........(3)$

$\bf \therefore \: 3(S_8 - S_4) = 3(8a + 28d - 4a - 6d) \\ = \bf3(4a + 22d) \\ \bf = 12a + 66d$

$\bf3(S_8 - S_4) = S_1_2$

$\bf \: Hence \: Proved \: .$