Question

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

Answer 1

Answer:

given,

a-2d+a-d+a+d+a+2d=2 or. 4a=2 or a=1/2. now a3+a4=11 given, so 1/2+2d+1/2+3d=11 or 5d=10 or d=2 so the terms are 1/2, 5/2, 9/2 and 13/2.

Answer 2

Step-by-step explanation:

sum of 4 terms = 2

sum of 3rd and 4th term= 11

$sn = \frac{n}{2} (2a + (n - 1)d)\\ s4 = \frac{4}{2} (2a + 3d) = 2 \\ 2(2a + 3d) = 2 \\ 2a + 3d = \frac{2}{2} = 1 \: \: \: \: \: equ(1)\\ \\ sum \: of \: 3rd \: and \: 4th \: term \\ a + 2d + a + 3d = 11 \\ 2a + 5d = 11 \: \: \: \: \: \: \: equ \: (2) \\equ \: (2) - (1) \\ 2d = 10 \\ d = \frac{10}{2} = 5$

$2a + 5d = 11 \\ 2a + 5(5) = 11 \\2a + 25 = 11 \\ 2a = 11 - 25 \\ 2a = - 14 \\ a = \frac{ - 14}{2} = - 7$

the terms are

a = -7

a+d= -7+5 = -2

a+2d=-7+10= 3

a+3d= -7+15=8

-7, -2, 3, 8

hope you get your answer