Question

Find the middle term of the sequence formed by all three-digit numbers which leave a remainder when divided by 7. Also find the sum of all numbers on both sides of the middle terms separately.

Answer 1

Hint: The sequence in the question is an arithmetic progression. The middle term of an arithmetic progression is ${{\left( \dfrac{n+1}{2} \right)}^{th}}$ term and the sum of the first n terms of an arithmetic progression is given by the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$ .

Complete step-by-step solution -
The sequence formed by all three-digit numbers which leave a remainder when divided by 7 is given below.
103, 110, 117,…………………………………………999 is an arithmetic progression.
In this arithmetic progression, first term = a = 103, common difference = d = 110-103 = 7 and the ${{n}^{th}}$ term$={{t}_{n}}=999$. Where n is the number of terms
We know that, the ${{n}^{th}}$ term of an arithmetic progression is given by
${{t}_{n}}=a+(n-1)d$
Now put the values of a, d and ${{t}_{n}}$, we get
$999=103+(n-1)7$
$999=103+7n-7$
$999=103-7+7n$

$999=96+7n$

$7n=999-96$

$7n=903$

Dividing both sides by 7, we get

n = 129.

The middle term of an arithmetic progression is given by
Middle term $={{\left( \dfrac{n+1}{2} \right)}^{th}}$term
Middle term $={{\left( \dfrac{129+1}{2} \right)}^{th}}$term
Middle term $={{\left( \dfrac{130}{2} \right)}^{th}}$term
Middle term $={{65}^{th}}$ term

Again applying the ${{n}^{th}}$ term formula for n =65, then the middle term is given by

${{t}_{65}}=103+(65-1)7=103+448=551$

Hence the middle term of an arithmetic progression is 551.

The sum of the first n terms of an arithmetic progression is given by
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$
There are 64 terms before the middle terms. So the sum of the first 64 terms of the given sequence is given by
${{S}_{64}}=\dfrac{64}{2}\left[ 2(103)+(64-1)7 \right]=32\left[ (206)+(63)7 \right]=32\left[ 206+441 \right]=32\times 647=20704$
Hence the sum of the terms before the middle term is 20704.

There are a total 129 terms in the sequence. So the sum of the first 129 terms of the given sequence is given by

${{S}_{129}}=\dfrac{129}{2}\left[ 2(103)+(129-1)7 \right]=\dfrac{129}{2}\left[ (206)+(128)7 \right]=\dfrac{129}{2}\left[ 206+896 \right]=\dfrac{129}{2}\times 1102=71079$

The sum of the terms after the middle terms = the sum of the total 129 term – the sum of the terms before the middle term –middle term

The sum of the terms after the middle terms $=71079-20704-551=49824$

Hence the sum of the terms after the middle terms is 49824.

Note: If a is the first term and d is the common difference of the AP having n terms, then ${{m}^{th}}$ term from the end is ${{(n-m+1)}^{th}}$ term form beginning. Here we had to form a series of 3-digit numbers which give the same remainder after dividing by 7.