Question

Find the sum of all numbers from 50 to 350 which are divisible by 4. and also find t15

Answer 1

Smallest number from 50 to 350 which is divisible by 4 = 52

Largest number from 50 to 350 which is divisible by 4 = 348

As 52 is the smallest number, it is the first term of the given arithmetic progression.

∴ T₁ = a = 52

As 348 is the largest number which is divisible by 4 between 52 and 350, it the last term of the given arithmetic progression.

We have to find the sum of terms which are divisible by 4, so common difference between the terms is 4.

T${}_{l} = a + ( l - 1 )d$

Where T${}_{l}$ is the last term and l is the number of terms.

We know, a${}_{n}=$a + ( n - 1 )d

∴ 348 = 52 + ( n - 1 )4

⇒ 348 - 52 = ( n - 1 )4

⇒ 296 = ( n - 1 )4

⇒ $\bold{\dfrac{296}{4} = n - 1 }$

⇒ 74 = n - 1

⇒ 74 + 1 = n

⇒ 75 = n

∴ Number of terms is 75.

∴ T₁₅ = a + ( 15 - 1 )d

        = 52 + ( 14 x 4 )  

        = 52 + 56

        = 108

15th term of the given arithmetic progression is 108.

From the properties of arithmetic sequence, we know that $S_{n} = [tex]\dfrac{n}{2}\bigg[2a + ( n -1 )d\bigg]$

Where S${}_{n}$ is the sum of n terms.

Substituting the values from the question in the formula,

⇒ $S_{75} = \dfrac{75}{2} \bigg[ 2( 52 ) + ( 75 - 1 )4 \bigg]$

⇒ $S_{75} = \dfrac{75}{2} \bigg[ 104 + 296 \bigg]$

⇒ $S_{75} = \dfrac{75}{2} \times 400$

⇒ $S_{75} = 75 x 200$

⇒ $S_{75} = 15000$

Therefore the sum of 75 terms is 15000 and 15th term is 108.