Question

in the arithmetic sequence 6,10,14 which term is 32 more than the 20th term​

Answer 1

28th is 32 more than the 20th term

Step-by-step explanation:

AP → 6, 10, 14

• a = 6
• d = 4

A_n = a + (n-1)d

A_20 = 6 + (20-1)4

= 6 + 19 × 4

= 82

We need to find the term which is 32 more than A_20 i.e

• A_20 + 32
• 82 + 32
• 114

A_n = a + (n-1)d

114 = 6 + (n - 1)4

114 - 6 = (n - 1)4

4(n - 1) = 108

n - 1 = 27

n = 28