Question

the sequence of three digit numbers below 500 which leaves remainder 1 when divided by 7.
find the no. of therms in the sequence.
sum of terms in the sequence ​

Answer 1

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

The series of three digit natural number less than 500 which leaves remainder 1 when divided by 7 is

$\rm :\longmapsto\:106, \: 113, \: 120, - - - - ,498$

So,

This sequence of numbers forms Arithmetic Progression with

↝ First term, a = 106

↝ Common difference, d = 113 - 106 = 7

↝ Last term, aₙ = 498

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,

$\rm :\longmapsto\:498 = 106 + (n - 1)7$

$\rm :\longmapsto\:498 - 106 = (n - 1)7$

$\rm :\longmapsto\:392 = (n - 1)7$

$\rm :\longmapsto\:56 = (n - 1)$

$\bf\implies \:n \: = \: 57$

• So, number of terms in sequence = 57

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic sequence is,

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

On substituting the values, we get

$\rm :\longmapsto\:S_n \: = \: \dfrac{57}{2}(106 + 498)$

$\rm :\longmapsto\:S_n \: = \: \dfrac{57}{2} \times 604$

$\rm :\longmapsto\:S_n \: = \: 57 \times 302$

$\bf\implies \:S_n = 17214$

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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.