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Let a be the first term and d be the common difference of the given AP. Then,
Sn= {tex}\frac{n}{2}{/tex}{tex} \cdot {/tex}[2a+(n-l)d],
{tex}\therefore{/tex} {tex}3(S_8-S_4) = 3{/tex}[{tex}\frac{8}{2}{/tex}{tex}(2a+7d)-{/tex}{tex}\frac{4}{2}{/tex}{tex}(2a+3d)]{/tex}
= {tex}3[4(2a+7d)- 2(2a+3d)] = 6(2a+11d){/tex}
{tex}= \frac { 12 } { 2 } \cdot ( 2 a + 11 d ) = S _ { 12 }{/tex}.
Hence, S12= 3(S8-S4).
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{tex}a_1 = 3x + y {/tex}and {tex}d = x - y{/tex}
{tex}a_2 = a_1 + d = 3x + y + x - y = 4x{/tex}
{tex}a_3 = a_2 + d = 4x + x - y = 5x -y{/tex}
{tex}a_4 = a_3 + d = 5x - y + x - y = 6x -2y{/tex}
So, the four terms are {tex} 3x + y, 4x, 5x - y {/tex} and{tex} 6x - 2y{/tex}.
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Gaurav Seth 6 years, 11 months ago
Let the vertices of triangle be A(1, -1), B(-4, 6) and C(-3, -5).
Then we have
x1 = 1, y1 = -1
x2 = -4, y2= 6
and x3= -3, y3 = -5
2Thank You