Question

The sum of four numbers in AP is 40 and the ratio of of the product of extremes to the product of means is 2:3. Find the numbers ?

Answer 1

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

${\boxed{\sf\:{First\;term\;be\;p}}}$

Also

${\boxed{\sf\:{Common\;difference\;be\;d}}}$

Hence,

Sum of four number of AP

p - 3d + p - d + p + d + p + 3d = 40

4p = 40

$\tt{\rightarrow p=\dfrac{40}{4}}$

p = 10

Now,

$\tt{\rightarrow\dfrac{(p+3d)(p+3d)}{(p-d)(p+d)}=\dfrac{2}{3}}$

$\tt{\rightarrow\dfrac{p^2-(3d)^2}{p^2-d^2}=\dfrac{2}{3}}$

$\tt{\rightarrow\dfrac{10^2-9d^2}{10^2-d^2}=\dfrac{2}{3}}$

$\tt{\rightarrow\dfrac{100-9d^2}{100-d^2}=\dfrac{2}{3}}$

3(100 - 9d²) = 2(100 - d²)

300 - 27d² = 200 - 2d²

100 = -2d² + 27d²

100 = 25d²

$\tt{\rightarrow\dfrac{100}{25}=d^2}$

4 = d²

√4 = d

±2 = d

$\large{\boxed{\sf\:{Substitute\;the\;values:-}}}$

${\boxed{\sf\:{For\;d=2}}}$

p - 3d = 10 - 3(2) = 10 - 6 = 4

p - d = 10 - 2 = 8

p + d = 10 + 2 = 12

p + 3d = 10 + 3(2) = 10 + 6 = 16

${\boxed{\sf\:{For\;d=-2}}}$

p - 3d = 10 - 3(-2) = 10 + 6 = 16

p - d = 10 - (-2) = 10 + 2 = 12

p + d = 10 + (-2) = 8

p + 3d = 10 + 3(-2) = 10 - 6 = 4

$\Large{\boxed{\sf\:{4,8,12,16\;or\;16,12,8,4}}}$

Answer 2

Answer:

${\boxed{\bold{4,8,12,16}}}$

$\tt Or,$ ${\boxed{\bold{16,12,8,4}}}$

Step-by-step explanation:

Let the first term be = P

Common difference be = d

Sum of four numbers in AP = 40

Now,

$\rightarrow\rm (p - 3d)+ (p - d)+ (p + d) + (p + 3d) = 40\\\rightarrow\rm 4p = 40 \\\rightarrow\rm p = \frac{40}{4} \\\rightarrow\rm p = 10$

Again,

$\rm \frac{(p+3d)(p-3d)}{(p+d)(p-d)} =\frac{2}{3} \\\rightarrow\rm \frac{{p}^{2}-9{d}^{2}}{{p}^{2}-{d}^{2}}=\frac{2}{3} \\\rightarrow\rm 3{p}^{2} - 27{d}^{2} = 2{p}^{2} - 2{d}^{2} \\\rightarrow\rm {p}^{2} = 25{d}^{2} \\\rightarrow\rm {10}^{2} = 25{d}^{2} \\\rightarrow\rm 100 = 25{d}^{2} \\\rightarrow\rm {d}^{2} =\frac{100}{25} \\\rightarrow\rm d =\pm \sqrt{4} \\\rightarrow{\boxed{\rm{d =\pm 2 }}}$

$\tt For\: d = 2 \: :-$

$\rm p - 3d = 10 - 3(2) = 4 \\\rm p - d = 10 - 2 = 8 \\\rm p + d = 10 + 2 = 12 \\\rm p + 3d = 10+3(2) = 16 \\\\\rightarrow{\boxed{\rm{Numbers = 4,8,12,16}}}$

$\tt For\: d = - 2\: :-$

$\rm p - 3d = 10-3(-2) = 16 \\\rm p-d = 10 - (-2)= 12 \\\rm p + d = 10+(-2) = 8 \\\rm p + 3d = 10+3(-2) = 4 \\\\\rightarrow{\boxed{\rm{Numbers=16,12,8,4}}}$