.
Find the sum of first 16 terms of an Arithmetic Progression whose
4th and 9th terms are - 15 and - 30 respectively.
Answers
Answer:
Answer:
-456
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.
The sum of 'n' terms of an AP is given by;
\dashrightarrow \ \sf S_{n} = \dfrac{n}{2} \Big\{2a + (n - 1)d\Big\}⇢ S
n
=
2
n
{2a+(n−1)d}
Where;
n = number of terms.
a = first term of the AP.
d = common difference of the AP.
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.
ATQ;
⇒ 4th term of the AP = -15
⇒ a₄ = -15
Using a\sf _n
n
= a + (n - 1)d we get;
⇒ a + (4 - 1)d = -15
⇒ a + 3d = -15 ⇒ Eq(1)
Also ATQ;
⇒ 9th term of the AP = -30
⇒ a₉ = -30
Using a\sf _n
n
= a + (n - 1)d we get;
⇒ a + (9 - 1)d = -30
⇒ a + 8d = -30 ⇒ Eq(2)
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
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]
From Equation 1;
⇒ a + 3d = -15
Substitute d = -3 above.
⇒ a + 3(-3) = -15
⇒ a - 9 = -15
⇒ a = -15 + 9
⇒ a = -6
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.
\dashrightarrow \ \sf S_{n} = \dfrac{n}{2} \bigg\{2a + (n - 1)d\bigg\}⇢ S
n
=
2
n
{2a+(n−1)d}
Where;
n = 16
a = -6
d = -3
\dashrightarrow \ \sf S_{16} = \dfrac{16}{2} \bigg\{2(-6) + (16 - 1)(-3)\bigg\}⇢ S
16
=
2
16
{2(−6)+(16−1)(−3)}
\dashrightarrow \ \sf S_{n} = 8 \ \bigg\{-12 + (15)(-3)\bigg\}⇢ S
n
=8 {−12+(15)(−3)}
\dashrightarrow \ \sf S_{n} = 8 \ \bigg\{-12 + (-45)\bigg\}⇢ S
n
=8 {−12+(−45)}
\dashrightarrow \ \sf S_{n} = 8 \ \bigg\{-12 - 45\bigg\}⇢ S
n
=8 {−12−45}
\dashrightarrow \ \sf S_{n} = 8 \ \bigg\{-57\bigg\}⇢ S
n
=8 {−57}
\dashrightarrow \ \sf S_{n} = -456⇢ S
n
=−456
∴ The sum of the first 16 terms of the given AP is -456.