Question

in an ap sum of first four trems 26 and first and 4th terms square sum of 125 find 4 trems of ap​

Answer 1

Step-by-step explanation:

(a) + (a + d) + (a + 2d) + (a + 3d) = 26

a + a + d + a + 2d + a + 3d = 26

4a + 6d = 26

2 (2a + 3d) = 26

2a + 3d = 26/2

2a + 3d = 13

3d = 13 - 2a

d = (13 - 2a)/3 ----------- {1}

(a)^2 + (a + 3d )^2 = 125

a^2 + [a + 3 × (13 - 2a) ÷ 3]^2 = 125

a^2 + [a + 13 - 2a]^2 = 125

a^2 + (13 - a)^2 = 125

a^2 + 13^2 - 26a + a^2 = 125

2a^2 - 26a + 169 = 125

2a^2 - 26a + 169 - 125 = 0

2a^2 - 26a + 44 = 0

2(a^2 - 13a + 22) = 0

a^2 - 13a + 22 = 0

(a^2 - 2a) + (- 11a + 22) = 0

a(a - 2) - 11 (a - 2) = 0

(a - 2) (a - 11) = 0

a - 2 = 0

a = 2

& a - 11 = 0

a = 11

From {1} : -

d = (13 - 2a)/3

d = [13 - 2(2)]/3

d = (13 - 4)/3

d = 9/3

d = 3

& d = [13 - 2(11)]/3

d = (13 - 22)/3

d = (-9)/3

d = -3

So, 1st AP = 2, 5, 8, 11.

So, 1st AP = 2, 5, 8, 11.And, 2nd AP = 11, 8, 5, 2.

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