Question

consider the number which are the multiple of 7 in between 100 and 400 ?find first and last term​

Answer 1

Answer:

Hope this helps.

Step-by-step explanation:

The first multiple of 7 that is greater than 100, is 105. (70+35 = 105).

The one before that is 98, and that's less than 100.

This is the most reliable way to solve the question (listing).

Another is to subtract 100 from 150 and divide that by 7 but that does not always work well. In this case, it does since 50 divided by 7 is 7.

Answer 2

Answer:

First let's find out the first number that is divisible by 7 in 100−500 .

15×7=105.

So, 105 is the first number.

Now, let's find the last number.

71×7=497 .

So, 497 is the last number.

The first number is 105 , let's assume that's our first number of a series(Arithmetic Series) and last is 497 . Common difference is 7 .

Now, nth term will be T(n)=105+(n−1)7

Here, solve for ′n′ by putting nth term as 497 .

497=105+(n−1)7

(n−1)7=497−105

(n−1)7=392

(n−1)=392/7

(n−1)=56

n=56+1

n=57

So, there are 57 terms in the series and hence there are 57 multiples of 7 in 100−500 .

Hope it helped.

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