Question

The sum of first 9 terms and the sum of next 8 terms of an arthemetic sequence are equal. If its common difference is 5
a) How many times of the common difference will be the difference between 10th and first term of this ap
b) what is the 9th term of this sequence.​

Answer 1

The sum of the nth terms of an AP can be given by:

$\Longrightarrow \sf S_n = \dfrac{n}{2} \Big(2a + (n-1)d\Big)$

ATQ

⇒ Sum of first 9th terms = Sum of the next 8 terms.

The sum of the next 8 terms of an AP can be written as;

Sum of next 8 terms = Sum of the first 17 terms - Sum of the first 9 terms.

S₈ = S₁₇ - S₉

Therefore;

$\Longrightarrow \sf S_9 = S_{17} - S_9$

$\Longrightarrow \sf S_9 + S_9= S_{17}$

$\Longrightarrow \sf 2S_9 = S_{17}$

$\Longrightarrow \sf 2\Bigg( \dfrac{n_{1}}{2} \Big(2a + (n_{1}-1)d\Big) \Bigg) = \dfrac{n_{2}}{2} \Bigg(2a + (n_{2}-1)d\Bigg)$

$\Longrightarrow \sf n_{1} \Big(2a + (n_{1}-1)d\Big) = \dfrac{n_{2}}{2} \Big(2a + (n_{2}-1)d\Big)$

$\Longrightarrow \sf 9 \Big(2a + (9-1)d\Big) = \dfrac{17}{2} \Big(2a + (17-1)d\Big)$

$\Longrightarrow \sf 9 \Big(2a + 8d\Big) = \dfrac{17}{2} \Big(2a + 16d\Big)$

$\Longrightarrow \sf 18a + 72d = \dfrac{17}{2} \times 2\Big(a + 8d\Big)$

$\Longrightarrow \sf 18a + 72d = 17\Big(a + 8d\Big)$

$\Longrightarrow \sf 18a + 72d = 17a + 136d$

$\Longrightarrow \sf 18d - 17a = 136d - 72d$

$\Longrightarrow \sf a = 64d$

$\Longrightarrow \sf a = 64(5)$

$\Longrightarrow \sf a = 320$

∴ The first term of the AP is 320.

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Part (a): How many times of the common difference will be the difference between 10th and the first term of this AP.

a₁ = 320

a₁₀ = ?

First, we find the 10th term.

⇒ a₁₀ = a + (n - 1)d

⇒ a₁₀ - a₁ =  (10 - 1)5

⇒ a₁₀ - a₁ = (9)5

⇒ a₁₀ - a₁ = (9)d

⇒ a₁₀ - a₁ = 9d

∴ The difference between the 10th & the 1st term of the AP is 9 times the original common difference.

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Part (b): What is the 9th term of this sequence?

⇒ a₉ = a + (n - 1)d

⇒ a₉ = 320 + (9 - 1)(5)

⇒ a₉ = 320 + (8)(5)

⇒ a₉ = 320 + 40

⇒ a₉ = 360

∴ The 9th term of the given AP is 360.