Question

Find the sum of the integers between 1 to 201 which are multiple of 4.

Answer 1

Let us solve this problem step by step.

On observation, we can conclude that the question asks us to find the sum of a specified 'n' terms in an arithmetic progression. First, let us find out the first and last term of this AP.
It is specified that the integers lie between 1 and 201, and are multiples of 4. Therefore, the first multiple of 4 is 4 itself and the last one in this range is 200 (as 4 x 50 = 200).  

We also need to find the no. of terms in this AP. For this, we have a readymade formula:
Tn = a + (n-1) d

where Tn = last term
a = first term
n = no. of terms
d = common difference

We know that the first term is 4, the last one is 200 and the common difference is 4, as all the numbers are multiples of 4.

Thus,
200 = 4 + (n-1)*4
200 - 4 = (n-1)*4
196 = (n-1)*4
196/4 = n-1
49 = n - 1
n = 49 + 1
n=50
Therefore, there are 50 terms in this AP.

Now, to find the sum to 'n' terms, we have the formula:
Sn = $\frac{n}{2}$ [ 2a + (n-1) d ]

Sn = $\frac{50}{2}$ [ 2*4 + (50-1)*4 ]

Sn = 25 (8 + 49*4)
Sn = 25 (8 + 196)
Sn = 25 (204)
Sn = 5100

Therefore, the sum of the integers between 1 to 201 which are multiple of 4 is equal to 5100.

Answer 2

The sum of the integers between 1 to 201 which are multiple of 4 = 5100

Step-by-step explanation:

The integers between 1 to 201 which are multiple of 4 are:

4, 8, 12, ................ ,200

The given sequence are in AP.

Here, first term (a) = 4, common difference (d) = 8 - 4 = 4 and

$a_{n}$ = 200

Let n be the number of terms.

To find, the sum of the integers between 1 to 201 which are multiple of 4 = ?

We know that,

The nth term of an AP

$a_{n}$ = a + (n - 1)d

⇒  4 + (n - 1)4 = 200

⇒ (n - 1)4 = 200 - 4 = 196

⇒ n - 1 = $\dfrac{196}{4}$ = 49

⇒ n = 49 + 1 = 50

The sum of up to nth term of an AP

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

= $\dfrac{50}{2} (4+200)$

= 25 × 204

= 5100

∴ The sum of the integers between 1 to 201 which are multiple of 4 = 5100