Question

the second term of an arithmetic sequence is 18 and its fourth term is 32 write its third and first terms

Answer 1

a(nth term)=a+(n-1)*d

So,

second term is-> a(2nd term)->a+d=18......................eq(i)  

(as,  a+(2-1)*d=a+1*d=a+d)

and, a(3rd term)-> a+3*d=32 .....................eq(ii)

(as, a+(4-1)d =a+3*d)

Now, eq(ii)-eq(i)

a+3*d-a-d=32-18

2*d=14

d=7

Putting value of d in equation (i)

a+d=18=>a+7=>18=>a=11

Hence, the first term=a=11 and the third term is a+(3-1)*d=a+2*d=11+7*2=25

Answer 2

Answer: the second team is 18 and fourth is 32.

18+32=50

50÷2=25

third team is 25

and the difference between second and third team is 7 so, first team is 18-7=11

first team is 11.