We know that a_3 = 9a3=9 and a_7 = 31a7=31.
We also know that a_n = a_1 + (n-1)dan=a1+(n−1)d.
a_7 = a_1 + (7-1)da7=a1+(7−1)d
Equation (1): 31 = a_1 + 6d31=a1+6d
and
a_3 = a_1 + (3-1)da3=a1+(3−1)d
Equation (2): 9 = a_1 + 2d9=a1+2d
Subtract Equation (2) from Equation (1).
31-9 = 6d -2d31−9=6d−2d
22 = 4d22=4d
d = 22/4d=224
Equation (3): d= 11/2d=112
Substitute Equation (3) in Equation (1).
31 = a_1 + 6d31=a1+6d
31 = a_1 + 6×11/2 = a_1 + 3331=a1+6×112=a1+33
a_1 = 31-33 = -2a1=31−33=−2
So a_1 = -2a1=−2 and d = 11/2d=112.
a_n = a_1 + (n-1)dan=a1+(n−1)d
So the 22nd term is given by
a_22 = -2 + (22-1)×11/2 = -2 + 21×11/2 = -4/2 + 231/2a22=−2+(22−1)×112=−2+21×112=−42+2312
a_22 = 227/2a22=2272
The sum S_nSn of the first nn terms of an arithmetic series is given by
S_n = (n(a_1+a_n))/2Sn=n(a1+an)2
So
S_22 = (22(a_1+a_22))/2 = 22/2×(-2+227/2) = 11×(-4/2 + 227/2) = 11 × 223/2S22=22(a1+a22)2=222×(−2+2272)=11×(−42+2272)=11×2232
S_22 = 2453/2S22=24532