Question

Find the sum of those integers numbers between 1and 500 divisible by 2 and 5

Answer 1

Answer: 12250

Hope this helps!

Step-by-step explanation: integers which are multiples of 10 are also the multiples of 2 and 5.

Therefore, multiples of 2 as well as of 5 between 1 and 500 are:

10, 20, 30, ...., 490 

Series forms an AP with first term,a=10 and common difference,d=20-10=10.

Let total number of terms of this AP be n.

Therefore, nth term of AP, an = Last term, l= 490 

an=a(n-1)d=l;

10+(n-1)10=490;

(n-1)10=480;

n-1=48;

n=48+1=49;

n=49

Thus,sum of n terms of AP is given as:

S₄₉=49/2 (10+490);

     =49/2 (500);

     =49×250=12250

Answer 2

Answer:-

• The sequence of numbers which are divisible by 2 between 1 and 500 is 2 , 4 , 6.....498.

• The sequence of numbers which are divisible by 5 between 1 and 500 is 5 , 10...495.

• The sequence of numbers which are divisible by both 2 and 5 between 1 and 500 is 10,20....490.

If we assume that this sequence is in AP,

• a (first term) = 10

• d(common difference) = 20 - 10 = 10.

• nth term (a(n)) = 490.

We know that,

nth term of an AP = a + (n - 1)d

$→ \sf{490 = 10 + (n - 1)(10)} \\ \\ →\sf{490 = 10 + 10n - 10} \\ \\→ \sf{490 = 10n} \\ \\ → \sf{n = \frac{490}{10} } \\ \\ →\sf \large{n = 49}$

Now,

Sum of the integers = Sum of the AP.

So,

We know that,

$\sf{Sum\:of\:n\:terms\:of\:an\:AP=\frac{n}{2}[2a+(n-1)d]}$

$→ \sf{s_{n} = \frac{49}{2} [2(10) + (49 - 1)10]} \\ \\ → \sf{ s_{n} = \frac{49}{2} (20 + 480)} \\ \\ → \sf{s _{n} = \frac{49}{2} (500)} \\ \\ →\sf \large{s _{n} = 12250}$

Hence, the sum of the integers which are divisible by both 2 and 5 between 1 and 500 is 12,250.