(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