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

Given :-

If 21th term and 30th term of Arthematic progression are 121 and 181 respectively

To Find :-

AP

Solution :-

n th term of AP is given by

$\sf a_n = a+(n-1)d$

Where

a = first term

n = given term

d = common difference

$\sf a_{21} = a+(21-1)d$

$\sf a_{21} = a + 20 \times d$

$\sf a_{21} = a+20d$

For 30 th term

$\sf a_{30} = a+(30-1)d$

$\sf a_{30} = a + 29\times d$

$\sf a_{30} = a+29d$

Subtracting them

$\sf a + 29d - a + 20d = 181-121$

$\sf 29d-20d=60$

$\sf 9d=60$

$\sf\dfrac{60}{9}=d$

$\sf\dfrac{20}{3}=d$

By putting value of d in eqn. 2

$\sf a + 29\times\dfrac{20}{3}=181$

$\sf a + \dfrac{580}{3} = 181$

$\sf a = \dfrac{181(3)+580}{3}$

$\sf a = \dfrac{-545+580}{3}$

$\sf a = \dfrac{-37}{3}$

40 th term

$\sf a = \dfrac{-37}{3} + 39\times \dfrac{20}{3}$

$\sf a = \dfrac{-37}{3}+13\times 20$

$\sf a =\dfrac{-37}{3} + 260$

$\sf a= \dfrac{-37 + 260(3)}{3}$

$\sf a = \dfrac{-37+780}{3}$

$\sf a = \dfrac{743}{3}$