Attempt of the following sub questions:
Find the sum of all numbers from 50 to 250 which divisible by 6 and find t₁₃.

Numbers from 50 to 250 which are divisible by 6 are

54, 60, 66, 72… 246

the series is in AP with common difference d = 6 and first term

a = 54

n^th term is 246

n^th term of A.P is given by t_n = a + (n – 1) × d

⇒ 246 = 54 + (n – 1) × 6

⇒ 246 = 54 + 6n – 6

⇒ 246 = 48 + 6n

⇒ 246 – 48 = 6n

⇒ 198 = 6n

∴ n = 33

Which means there are 33 numbers between 50 and 250 which are divisible by 6

Sum of an AP is given by × (2a + (n – 1) × d)

⇒ sum = × ((2 × 54) + (33 – 1) × 6)

⇒ sum = × (108 + 32 × 6)

⇒ sum = × (108 + 192)

⇒ sum = × (108 + 192)

⇒ sum = × (300)

⇒ sum = 33 × 150

∴ sum = 4950

13^th term using equation for t_n

⇒ t₁₃ = 54 + (13 – 1) × 6

⇒ t₁₃ = 54 + (12) × 6

⇒ t₁₃ = 54 + 72

⇒ t₁₃ = 126

Therefore, sum of all numbers from 50 to 250 which divisible by 6 is 4950 and 13^th term t₁₃ is 126