Question

if the 6th and 13th term of arithmetic progression are 22 and 43 respectively then what is the sum of the first 21 terms?

Answer 1

Solution

Given :-

• 6th terms of A.P. = 22
• 13th terms of A.P. = 43

Find :-

• Sum of the 21 terms

Explanation

We using here formula ,

★ nth terms = a + (n - 1)d

Where,

• a = first terms
• n = total number of terms
• d = common Defference

According to question,

⇒6th terms = a + (6 - 1)d

⇒ 22 = a + 5d ____________________(1)

And,

⇒ 13th terms = a + (13 - 1)d

⇒ 43 = a + 12d ___________________(2)

Sub. equ(1) & equ(2)

⇒ 5d - 12d = 22 - 43

⇒-7d = -21

⇒d = -21/(-7)

⇒d = 3

Now, keep value of d in equ(1)

we get,

⇒22 = a + 5×3

⇒a = 22 - 15

⇒ a = 7

Now, we have

• a(first terms) = 7
• d ( common Defference) = 3

Now, we calculate second term

For this , keep value in above formula

⇒ 2nd terms = 7 + (2 - 1)3

⇒ 2nd terms = 7 + 3

⇒ 2nd terms = 10

Again,

⇒ 3rd terms = 7 + (3 - 1)3

⇒ 3rd terms = 7 + 3 × 2

⇒ 3rd terms = 7 + 6

⇒ 3rd terms = 13

Again,

⇒ 4rd terms = 7 + (4 - 1)3

⇒ 4rd terms = 7 + 3 × 3

⇒ 4rd terms = 7 + 9

⇒ 4rd terms = 16

similarly ,

⇒ 5th terms = 19

And,

⇒ 6th terms = 22

Since, series will be

• 7 , 9 , 11 , 13, 16 , 19 , 22 .______ up to 21terms .

Now, we calculate sum of all 21 terms .

Using Formula

★ Sn = n/2 [ 2a + (n - 1)d]

Where,

• Sn = sum of nth terms .
• n = Number of terms
• a = first terms
• d = common Defference

Now, keep all required values

⇒S21 = 21/2 [ 2×7 + (21-1)3]

⇒S21 = 21/2 [ 14 + 20 × 3]

⇒S21 = 21/2 [ 14 + 60]

⇒S21 = (21/2) × 74

⇒S21 = 21 × 37

⇒S21 = 777

Hence

• Sum of 21th terms will be = 777

___________________

Answer 2

Answer:

777

please mark me as brainliest