Question

The sum of four consecutive terms of an ap is 2. the sum of the third and fourth term is 11​

Answer 1

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Answer 2

$\huge\colorbox{pink}{Answer ✔}[/tx]</p><p></p><p>[tex]sum of 4 terms = 2</p><p></p><p>sum of 3rd and 4th term= 11</p><p></p><p>\begin{gathered} sn = \frac{n}{2} (2a + (n - 1)d)\\ s4 = \frac{4}{2} (2a + 3d) = 2 \\ 2(2a + 3d) = 2 \\ 2a + 3d = \frac{2}{2} = 1 \: \: \: \: \: equ(1)\\ \\ sum \: of \: 3rd \: and \: 4th \: term \\ a + 2d + a + 3d = 11 \\ 2a + 5d = 11 \: \: \: \: \: \: \: equ \: (2) \\equ \: (2) - (1) \\ 2d = 10 \\ d = \frac{10}{2} = 5\end{gathered}sn=2n(2a+(n−1)d)s4=24(2a+3d)=22(2a+3d)=22a+3d=22=1equ(1)sumof3rdand4thterma+2d+a+3d=112a+5d=11equ(2)equ(2)−(1)2d=10d=210=5</p><p></p><p>\begin{gathered}2a + 5d = 11 \\ 2a + 5(5) = 11 \\2a + 25 = 11 \\ 2a = 11 - 25 \\ 2a = - 14 \\ a = \frac{ - 14}{2} = - 7\end{gathered}2a+5d=112a+5(5)=112a+25=112a=11−252a=−14a=2−14=−7</p><p></p><p>the terms are</p><p></p><p>a = -7</p><p></p><p>a+d= -7+5 = -2</p><p></p><p>a+2d=-7+10= 3</p><p></p><p>a+3d= -7+15=8</p><p></p><p>-7, -2, 3, 8</p><p></p><p>$

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