Question

Four parts of 24 are in A.P. such that the product of extremes is to product of means is $7:15$ then four parts are

• A. $\frac{3}{2},\frac{9}{2},\frac{15}{2},\frac{21}{2}$
• B. $\frac{11}{2},\frac{13}{2},9$
• C. $\frac{5}{2},\frac{15}{2},\frac{9}{2},\frac{21}{2}$
• D. $\frac{21}{2},\frac{9}{2},\frac{15}{2},\frac{5}{2}$

Answer 1

The correct option is A $\frac{3}{2},\frac{9}{2},\frac{15}{2},\frac{21}{2}$

As the number are even so consider them as (common difference $2d$)
$a-3d,a-d,a+d,a+3d$

According to problem $S_4=4a=24\Rightarrow a=6$

Again $\frac{(a-3d)(a+3d)}{(a-d)(a+d)}=\frac{7}{15}$

$\therefore d=\pm \frac{3}{2}$

Hence required A.P is $\frac{3}{2},\frac{9}{2},\frac{15}{2},\frac{21}{2}$