Math, asked by sh2424, 11 months ago

Hiiiiii buddies✌
pls tell y can't v let the numbers as a-d, a-2d etc and y we take it as a-3d ,a-d ,a+d
#pls dont answer I f u dont know the answer#​

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Answers

Answered by ShuchiRecites
6

Why 4 terms are taken are taken as a - 3d, a - d, a + d and a + 3d?

These are assumed values to be learnt however we can take the values even as a, a + d, a + 2d and a + 3d. But it depends upon use because we need to cancel out ‘d’ terms hence we take the first sequence.

Solution

Let terms be a - 3d, a - d, a + d and a + 3d.

→ a - 3d + a - d + a + d + a + 3d = 26

→ 4a = 26

a = 13/2

→ (a - 3d)² + (a - d)² + (a + d)² + (a - 3d)² = 214

→ a² + 9d² - 6d + a² + d² - 2ad + a² + d² + 2ad + a² + 9d² + 6d = 214

→ 4a² + 20d² = 214

→ 4(13/2)² + 20d² = 214

→ 169 + 20d² = 214

→ 20d² = 45

d = √(9/4) or ± 3/2

Formation of A.P.,

→ a - 3d = 13/2 - [± 3(3/2)] = 2 or 11

→ a - d = 13/2 - [± (3/2)] = 5 or 8

→ a + d = 13/2 + [± 3/2] = 8 or 5

→ a + 3d = 13/2 + [± 3(3/2)] = 11 or 2

Hence terms in A.P., are 11, 8, 5, 2 or 2, 5, 8, 11.

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