Question

find the sum of all natural numbers between 250 and 1000 which are correctly divisible by 8?​

Answer 1

Step-by-step explanation:

Nitin was a fourteen year old boy. after returning from school , he kept his school bag and ran to the backyard to play. suddenly he noticed a shadow and when he turned back .

Answer 2

$\large\underline{\sf{Solution-}}$

The natural numbers between 250 and 1000 which are divisible by 8 are as

$\rm :\longmapsto\:256, \: 264, \: 272, \: - - - - ,992$

So, its an AP series with

• First term, a = 256

• Common difference, d = 8

• nth term = 992

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

Tʜᴜs,

$\rm :\longmapsto\:992 = 256 + (n - 1)8$

$\rm :\longmapsto\:992 - 256 = (n - 1)8$

$\rm :\longmapsto\:736 = (n - 1)8$

$\rm :\longmapsto\:92 = n - 1$

$\bf\implies \:n \: = \: 93$

Now,

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(\:a\:+ \: a_n \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

aₙ is the nᵗʰ term.

Thus,

$\rm :\longmapsto\:S_n = \dfrac{93}{2} \bigg(256 + 992 \bigg)$

$\rm :\longmapsto\:S_n = \dfrac{93}{2} \bigg(1248 \bigg)$

$\rm :\longmapsto\:S_n = 93 \times 624$

$\bf\implies \:\boxed{\tt{ S_n = 58032}}$

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Additional Information

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.