Question

write three arithmetic sequence with 30 as the sum of the first five terms

Answer 1

As the number of terms is odd, let the middle (5+12=3rd)(5+12=3rd) term be aa

So, the 55 terms will be a,a±d,a±2da,a±d,a±2d

So, (a−2d)+(a−d)+a+(a+d)+(a+2d)=30⟹5a=30⟹a=6(a−2d)+(a−d)+a+(a+d)+(a+2d)=30⟹5a=30⟹a=6

and (a−2d)2+(a−d)2+a2+(a+d)2+(a+2d)2=220(a−2d)2+(a−d)2+a2+(a+d)2+(a+2d)2=220

⟹a2+2{a2+d2}+2{a2+(2d)2}=220⟹a2+2{a2+d2}+2{a2+(2d)2}=220 as (A+B)2+(A−B)2=2(A2+B2)(A+B)2+(A−B)2=2(A2+B2)

⟹10d2=220−5⋅62=40⟹d2=4⟹10d2=220−5⋅62=40⟹d2=4

Had the number of terms been even, we could take the terms as a±d,a±3d,⋯

Answer 2

Answer:

sum(5) = (5/2)(2a + 4d) = 30

2a + 4d = 12

a + 2d = 6

a = 6-2d

so we can assign any value to d

e.g. d = 7

then a = 6-14 = -8

one possible sequence is:

-8 , -1, 6, 13, 20

(their sum is 30)

let d = 3

a = 6-6 = 0

terms are:

0, 3, 6, 9, 12

(their sum is 30)