Question

The sum of three consecutive terms of an arthematic progression is 21 then find the terms

Answer 1

let three consecutive terms be = x,x+1,x+3

there sum will be 21

so by Sn formula

n/2(2a+(n-1)d)

d=x+1-x

d=1

Sn=21

n=3(because there are sum of three terms)

Sn=n/2(2a+(n_1)d

21=3/2(2a+(3_1)1

21×3=6a+3-3

42=6a

a=42/6

a=7

a2=a+d=6+1

=7

a3=2a+1

2×6+1

13

hope now you understand that☺️

Answer 2

Answer:

let three consecutive terms be = x,x+1,x+3

there sum will be 21

so by Sn formula

n/2(2a+(n-1)d)

d=x+1-x

d=1

Sn=21

n=3(because there are sum of three terms)

Sn=n/2(2a+(n_1)d

21=3/2(2a+(3_1)1

21×3=6a+3-3

42=6a

a=42/6

a=7

a2=a+d=6+1

=7

a3=2a+1

2×6+1

13

Step-by-step explanation:

@Genius