What is the range of a set consisting of the first 100 multi

\(a_1=77\)
\(a_{100}=77+(99)*7=693+77\)

Range = 693+77-77=693

Ans: "A"

A
the first 100 multiples of 7 is 7-700, so the range is 693

It does not matter if you start at 7 or 77

First Multiple greater than 70 -> 7 * 11 = 77
100th Multiple greater than 70 -> 7 * (11+100) = 770

Range is 770-77 = 693....

Pls can somebody explain where I am making a mistake and why is the answer 777 and not 693

There are a couple of mistakes you made but they canceled out to give the right answer! (693 is the right answer)

100th multiple greater than 70 will be 7 (10 + 100) or we can say 70 + 7*100 because it has to be the 100th multiple after 70.
When you do 7 * (11+100) i.e. 77 + 7*100, you are calculating the 101th multiple since you are finding the 100th multiple after 77, not 70.

The second error is that 7 * (11+100) = 777, not 770
Actually, 7 (10 + 100) = 770

The range is 770 - 77 = 693
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First multiple of 7 after 70= 77
100th multiple of 7 after 70= 110th multiple of 7= 770

Range= 100th multiple-first multiple= 770-77= 693.

Ans: A.

Ankitk, could you please check whether you've posted the correct OA? Thanks.
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sequence is as follows.

77,84,....770

Range = 770-77 = 693

Answer A.

100 multiples of 7 greater than 70 will be as follows :

77+84+91+...

= 7 ( 11 + 12 + 13 + .....+110 )

so the last term is 770.

Range = 770 - 77 = 693.

Ans : A

why is 99 is multiplied by 7?

In an arithmetic progression
\(A_n=A_1+(n-1)*d\)

{77,84,91,98,.....,}
\(A_1=First \hspace{2} term=77\)
\(d=Common \hspace{2} Difference=7\)

\(A_{100}=A_1+(n-1)*d\)
\(A_{100}=77+(100-1)*7\)
\(A_{100}=77+99*7\)
\(A_{100}-A_1=77+99*7-77=99*7\)

The solutions above are great - just another way to look at things (and the answer certainly is A here):

Multiples are equally spaced. If you take any set of 100 consecutive multiples of 7, the range will always be the same; the smallest number in the list is irrelevant. So you can ignore the 'greater than 70' condition in the question, and just look at the first 100 positive multiples of 7: {7, 14, 21, 28, ..., 700}, which has a range of 693.
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why is 99 is multiplied by 7?[/quote]

In an arithmetic progression
\(A_n=A_1+(n-1)*d\)

{77,84,91,98,.....,}
\(A_1=First \hspace{2} term=77\)
\(d=Common \hspace{2} Difference=7\)

\(A_{100}=A_1+(n-1)*d\)
\(A_{100}=77+(100-1)*7\)
\(A_{100}=77+99*7\)
\(A_{100}-A_1=77+99*7-77=99*7\)[/quote]

Hey fluke
i did it bit simpler

without any condition highest no among the first 100 multiple would be 100 x 7 =700
and highest among the first 10 multiple would be 10 x 7 = 70
so if we want to skip the first 10 multiple then we just need to add 70 to the previous 700 right?
it came 770 and now first multiple greater than 70 is simply 77
so answer is 770-77 = 693

please guide me if i am right

I got the answer as 693. Here is how I approached it.
first multiple of 7 after 70 is = 70 + 7.
second multiple of 7 after 70 = 70 + 7 + 7.
Here is a pattern 70 + n x 7.

If n =100 => we get 770.

range 770 - 77 = 693.

last term = d*(n-1)
d = 7, n=100
7(99) = 693

A.

Another logical approach:
In this question, we talk about only positive multiples.

First multiple of 7 = 7*1
Second multiple of 7 = 7*2
...
100th multiple of 7 = 7*100
But we don't want the first ten multiples (upto 7*10).
First acceptable multiple = 7*11 = 77
So to get a total of 100 multiples, we need to go 10 steps extra after 100 and take upto 110th multiple.
110th multiple of 7 = 7*110 = 770
Range = 770 - 77 = 693
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A

Solved it following way

number of elements in the set formula = (last multiple of x) - (first multiple of x) / x + 1

first multiple of x >70 = 77
(last multiple of x) = x
number of elements in the set = 100

note (x-77) is also a range of set

therefore
(x-77) / 7 + 1 = 100
(x-77) / 7 = 99
(x-77) = 99 * 7 = 693

\(a_1 = 77\)
\(a_2 = 84\)
\(a_3 = 91\)
\(a_n = 77 + (n-1)*7\)

Range:\(a_{100} - a_{1} = (n-1)*7\)

\(99*7 = 693\)

I select B first .. then i cheked the OA.... bt my mistake was i was subtracting 70 from 770.. but question is saying above 70..


but i did this way..

till 70 its 10 multiple of 7 but we dont need those multiples, so we will calculate 110 multiple of 7..

7*110=770..

question is saying multiple of 7 above 70.. so first multiple of 7 after 70 is 77 .. so so so.. to get the range we subtract max-min value..

770-77=693...Answer is A..
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First multiple greater than \(70: 7x10\) is \(7x11\). 100 first multiples so our last multiple is \(7x110\).

Range = Maximum - Minimum of a set S.

\((7x110) - (7x11) = 7x99 = 693\)

Guys, I used the digits approach, and it worked out

The first multiple of 7 greater than 70 = 77 --> last digit = 7
The 10th multiple of 7 greater than 70 would be = 140 --> last digit = 0
(note that every 10th multiple of 7 would end in 0) So the 100th multiple should also end in 0

So the range = xx0 - 77 = xx3 --> last digit = 3

Only option A satisfies!

Let me know if this can be an approach to solve.

thanks!