Question

.
Find the sum of first 16 terms of an Arithmetic Progression whose
4th and 9th terms are - 15 and - 30 respectively.​

Answer 1

Answer:

please mark it as brainliest

Answer 2

Answer:

-456

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Step-by-step explanation:

According to the question, we have an AP where the 4th term is -15 and the 9th term is -30, and we're asked to find the sum of the first 16 terms.

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The sum of 'n' terms of an AP is given by;

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

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Where;

• n = number of terms.

• a = first term of the AP.

• d = common difference of the AP.

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We know that n = 16, but we do not know the values of 'a' and 'd', we can use the data given in the question to find out their values.

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ATQ;

⇒ 4th term of the AP = -15

⇒ a₄ = -15

Using a$\sf _n$ = a + (n - 1)d we get;

⇒ a + (4 - 1)d = -15

⇒ a + 3d = -15 ⇒ Eq(1)

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Also ATQ;

⇒ 9th term of the AP = -30

⇒ a₉ = -30

Using a$\sf _n$ = a + (n - 1)d we get;

⇒ a + (9 - 1)d = -30

⇒ a + 8d = -30 ⇒ Eq(2)

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Subtracting Eq(1) and Eq(2) we get;

⇒ a + 3d - (a + 8d) = -15 - (-30)

⇒ a + 3d - a - 8d = -15 + 30

⇒ -5d = 15

⇒ d = -15/5

⇒ d = -3

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Now we've got the value of 'd' the common difference, let's substitute this value in Eq(1) to get the value of 'a'. [You can substitute it in Eq(2) as well, your choice]

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From Equation 1;

⇒ a + 3d = -15

Substitute d = -3 above.

⇒ a + 3(-3) = -15

⇒ a - 9 = -15

⇒ a = -15 + 9

⇒ a = -6

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Now we've got all the values we need to find the sum of the first 16 terms, so we'll substitute them in the formula to get the required result.

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$\dashrightarrow \ \sf S_{n} = \dfrac{n}{2} \bigg\{2a + (n - 1)d\bigg\}$

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Where;

• n = 16

• a = -6

• d = -3

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$\dashrightarrow \ \sf S_{16} = \dfrac{16}{2} \bigg\{2(-6) + (16 - 1)(-3)\bigg\}$

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$\dashrightarrow \ \sf S_{n} = 8 \ \bigg\{-12 + (15)(-3)\bigg\}$

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$\dashrightarrow \ \sf S_{n} = 8 \ \bigg\{-12 + (-45)\bigg\}$

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$\dashrightarrow \ \sf S_{n} = 8 \ \bigg\{-12 - 45\bigg\}$

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$\dashrightarrow \ \sf S_{n} = 8 \ \bigg\{-57\bigg\}$

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$\dashrightarrow \ \sf S_{n} = -456$

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∴ The sum of the first 16 terms of the given AP is -456.