Question

Find four consecutive terms in an AP whose sum is 12 and sum of 3rd and 4th term is 14
(Assume the four consecutive terms in AP are a-d, a, a+d, a+2d

Answer 1

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հίί ʍαtε_______✯◡✯
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һєяє' ʏȏȗя ѧṅśwєя ʟȏȏҡıṅɢ ғȏя________
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✯ Let the four consecutive terms in an A.P. be a – 3d, a – d, a + d, a + 3d


As per the first condition,


a – 3d + a – d + a + d + a + 3d = 12


∴  4a = 12


∴ a = 12/4


∴ a = 3 ...........eq. (i)


As per the second condition,


a + d + a + 3d = 14


∴ 2a + 4d = 14



∴ 2 (3) + 4d = 14 [From eq. (i)]


∴ 6 + 4d = 14


∴ 4d = 14 – 6


∴ 4d = 8


∴ d = 8/4


∴ d = 2


∴ a – 3d = 3 – 3 (2) = 3 – 6 = – 3


a – d = 3 – 2 = 1


a + d = 3 + 2 = 5


a + 3d = 3 + 3 (2) = 9

∴ The four consecutive terms of A.P. are – 3, 1, 5 and 9

$hope \: it \: helps$
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Answer 2

հίί ʍαtε_______✯◡✯

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_________________________

һєяє' ʏȏȗя ѧṅśwєя ʟȏȏҡıṅɢ ғȏя________

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✯ Let the four consecutive terms in an A.P. be a – 3d, a – d, a + d, a + 3d

As per the first condition,

a – 3d + a – d + a + d + a + 3d = 12

∴ 4a = 12

∴ a = 12/4

∴ a = 3 ...........eq. (i)

As per the second condition,

a + d + a + 3d = 14

∴ 2a + 4d = 14

∴ 2 (3) + 4d = 14 [From eq. (i)]

∴ 6 + 4d = 14

∴ 4d = 14 – 6

∴ 4d = 8

∴ d = 8/4

∴ d = 2

∴ a – 3d = 3 – 3 (2) = 3 – 6 = – 3

a – d = 3 – 2 = 1

a + d = 3 + 2 = 5

a + 3d = 3 + 3 (2) = 9

∴ The four consecutive terms of A.P. are – 3, 1, 5 and 9