(1) 7

(2) 9

(3) –7

(4) –27

**Ans. (3)** –7

**Sol.** $B(\bar{x})=a\bar{x}+b={a(1+2+3+.....+17)\over17}+b=17$

${a.(17.18)\over17.2}+b=17$

9a + b = 17 .........(i)

$\alpha A^2={\Sigma x^2\over n}-\left({\Sigma x\over n}\right)^2$

$={1^2+2^2+.....+17^2\over17}-\left({1+2+.....+17\over17}\right)^2$

$={17.18.35\over6.17}-\left({17.18\over2.17}\right)^2$

= 105 – 81 = 24

$\therefore\alpha^2_B=a^2\alpha_A^2=a^2.24=216$

$a^2={216\over24}=9$

a = 3

$\therefore$ b = 17 – 27

b = –10

$\therefore$ a + b = –7