पुस्तक विक्रेता ने वीपीपी द्वारा आपको पुस्तक भेजी है?? धन्यवाद देते हुए पत्र लिखिए।
कृपया करके कोई भी फालतू का उत्तर न दे कृपया करके मुझे सही उत्तर दीजिए कृपया!!
Answers
Answer:
Photosynthesis, the process by which green plants and certain other organisms transform light energy into chemical energy. During photosynthesis in green plants, light energy is captured and used to convert water, carbon dioxide, and minerals into oxygen and energy-rich organic compounds.
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.
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.