Question

Find the sum of all four digit numbers which are divisible by 7?

Answer 1

Answer:

7071071

Step-by-step explanation:

The smallest four digit number divisible by 7 is 1001

and the largest four digit number divisible by 7 is 9996

So, our series look like: 1001, 1008, 1015,...., 9996

The given sequence is Arithmetic Progression.

Arithmetic Progression is a sequence in which every two neighbor digits have equal distances.

For finding the nth term, we use formula

aₙ = a + (n - 1) d

where, aₙ = value of nth term

a = First term

n = number of term

d = difference

a = 1001, d = 7, aₙ = 9996

∴ 9996 = 1001 + (n - 1) × (7)

⇒ n = 1286

Now, Finding the sum of the given series.

For sum of Arithmetic Sequence we have a formula,

Sₙ = n÷2[2a + (n - 1) d ]

⇒ Sₙ = 1286÷2 [2 × 1001 + 1285 × 7 ]

⇒ Sₙ = 7071071

Answer 2

The sum of all four digit numbers which are divisible by 7 = 7071071

Step-by-step explanation:

The four digit numbers which are divisible by 7:

1001, 1008, 1015, ............, 9996

The given number are in AP.

Here, first term (a) = 1001, common difference (d) = 1008 - 1001 = 7

and last term (l) = 9996

Let n be the number of terms.

To find, the sum of all four digit numbers which are divisible by 7.

We know that,

l = a + (n - 1)d

⇒ 1001 + (n - 1)7 = 9996

⇒ (n - 1)7 = 9996 - 1001 = 8995

⇒ n - 1 = 1285

⇒ n = 1285 + 1 = 1286

The sum of all four digit numbers which are divisible by 7

= $\dfrac{n}{2} (a+l)$

= $\dfrac{1286}{2} (1001+9996)$

= 643( 10997)

= 7071071

∴ The sum of all four digit numbers which are divisible by 7 = 7071071