Find the sum of all integers wich are divisible by 7 and lying between 50 and 500.

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Numbers between 50 and 500

The first number dividible by 7 is 56 , then 56+7 , ......500

56, (56+7) 56+ (2*7) + (56+3*7) + .....+ (56 + 63*7)

Then the sum:

Sn = (n/2)*(an+a1)

     = 64/2 (497+ 56)

     = 32* 553

     = 17,696

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Let's verify which are the integers divisible by seven, located between 50 and 500.

Between 50 and 60 is the number 56

Between 60 and 70 is the number 63 and 70

Between 70 and 80 is 77

And so on...

Between 490 and 500 is 497

The sum we have to calculate is :

S = 56 + 63 + 70 + ...... + 497

We do not know the number of terms in this sum, but we notice that the terms of the sum is the terms of an arithmetical series, whose the first term is a1 = 56 and the common difference is d = 7.

The numbers of term os the sum is n and the last term, an  =497.

Let's apply the formula of the general termof an a.s.

an = a1 + (n-1)*d

497 = 56 + (n-1)*7

We'll remove the brackets:

497 = 56 + 7n - 7

We'll move like terms to the left side and we'll isolate n to the right side:

497-56+7 = 7n

448 = 7n

n = 64

So, the number of terms in the sum is 64.

S64 = (a1 + a64)*64/2

S64 = (56+497)*64/2

S64 = 553*32

S64 = 17696

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

a1 = 56= 7*8  and an = 497 = 71*7is the 1st  and last terms.multiple of 7 in 50 to 500.

So Sn = 56+63+70+.....497 = 7(8+9+10+....71),  number of terms = 71-8+1 = 64.

= 7 (8+71)(64/2) as the terms 8 to 71 are natural numbers whose sum = (1st term+last term)(number of terms/2)

= 7*79*64/2

=17696

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