Question

[Find the 11th term of an arithmetic sequence,
having 21 terms and the sum of first 21 terms
is 420.]

Answer 1

Answer:

a11 = 20

Step-by-step explanation:

Given : n = 21 and SUM 21 = 420

SUMn = n/2 × [2a + (n-1)d]

420 = 21/2 × [ 2a + (21 - 1)d]

420 = 21/2 × [ 2a + 20d]

420 = 21 [a + 10d]

420/21 = a + 10d

20 = a + 10d ---------(1)

a11 = a + (11-1)d

a11 = a + 10d ---------(2)

Eliminate both rhe equation,

20 = a + 10d

-a11 =- a -10d ( While eliminating the sign gets changed)

____________

20 - a11 = 0

____________

By eliminating we get , 20 - a11 = 0

Hence a11 = 20

I hope you will understand this solution easily.