Question

The sum of the first ten terms of a linear sequence is 145. the sum of the next ten term is 445. find the sum of the first four term of the sequence

Answer 1

The sum of the first four terms of the sequence is 22.

In this question,

The formula of sum of linear sequence is

$S_n =\frac{n}{2}(2a+(n-1)d)$

The sum of the first ten terms of a linear sequence is 145

⇒ $S_{10} =\frac{10}{2}(2a+(10-1)d)$

⇒ 145 = 5 (2a+9d)

⇒ $\frac{145}{5} =2a+9d$

⇒ 29 = 2a + 9d  ------- (1)

The sum of the next ten term is 445, so the sum of first twenty terms is

⇒ 145 + 445

⇒ $S_{20} =\frac{20}{2}(2a+(20-1)d)$

⇒ 590 = 10 (2a + 19d)

⇒ $\frac{590}{10}=2a+19d$

⇒ 59 = 2a + 19d -------- (2)

Now subtract (2) from (1),

⇒ 30 = 10d

⇒ d = $\frac{30}{10}$

⇒ d = 3

Substitute d in (1), we get

⇒ 29 = 2a + 9(3)

⇒ 29 = 2a + 27

⇒ 29 - 27 = 2a

⇒ 2 = 2a

⇒ a = $\frac{2}{2}$

⇒ a = 1

Thus, sum of first four terms is

⇒ $S_4 =\frac{4}{2}(2(1)+(4-1)(3))$

⇒ $S_4 =2(2+(3)(3))$

⇒ S₄ = 2(2+9)

⇒ S₄ = 2(11)

⇒ S₄ = 22.

Hence we can conclude that the sum of the first four terms of the sequence is 22.

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