Question

if 21th term and 30th term of Arthematic progression are 121 and 181 respectively find the 40th term of Arthematic progression.
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Answer 1

Answer:-

Given:-

21st term of an AP = 121

30th term of an AP = 181

We know that,

nth term of an AP (aₙ) = a + (n - 1)d

So,

⟹ a₂₁ = 121

⟹ a + (21 - 1)d = 121

⟹ a + 20d = 121 -- equation (1)

Similarly,

⟹ a₃₀ = 181

⟹ a + 29d = 181 -- equation (2)

Subtract equation (1) from (2).

⟹ a + 29d - (a + 20d) = 181 - 121

⟹ a + 29d - a - 20d = 60

⟹ 9d = 60

⟹ d = 60/9

⟹ d = 20/3

Substitute the value of d in equation (1).

⟹ a + 20(20/3) = 121

⟹ a + 400/3 = 121

⟹ a = 121 - 400/3

⟹ a = (363 - 400)/3

⟹ a = - 37/3

Now,

⟹ a₄₀ = a + 39d

⟹ a₄₀ = - 37/3 + 39(20/3)

⟹ a₄₀ = ( - 37 + 780)/3

⟹ a₄₀ = 743/3

∴ The 40th term of the given AP is 743/3.

Answer 2

Answer:

if 21th term and 30th term of Arthematic progression are 121 and 181 respectively find the 40th term of Arthematic progression.

if 21th term and 30th term of Arthematic progression are 121 and 181 respectively find the 40th term of Arthematic progression.

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