Question

#pls help..
please.....​

Answer 1

Answer:

first term = 5

sum of first 4 terms = (1/2)( sum of next 4 terms)

Sum of first 4 terms + sum of next 4 terms = sum of first 8 terms

sum of first 4 terms + 2(sum of first4 terms = sum of first 8 terms

(3)( sum of first 4 terms) = sum of first 8 terms

first term(a) = 5 ,let d be common difference

(3)( (4/2)( 2(5) +(4-1)d) = (8/2)( 2(5)+(8-1)d)

(3)( 2( 10+ 3d))= 4( 10 +7d)

(3)( 10+ 3d) = 2( 10+7d)

30+ 9d = 20+ 14d

10 =5d

d=2

Answer 2

Answer:

first term a=5

a1+a2+a3+a4=1/2 (a5+a6+a7+a8) (given)

a+(a+d)+(a+2d)+(a3d)=1/2 ((a+4d)+(a+5d)+(a+6d)+(a+7d)

(an=a+(n-1)×d)

4a+6d=1/2 (4a+22d)

8a+12d=4a+22d

4a=10d

d=2/5a

d=2/5×5(a=5)

d=2

hence the common difference of the above. p is