Question

find three such number which are in ap sum of their first and third term is 12 and product
of first and second term is 24
​

Answer 1

Solution :-

Let :-

First term = a

Common difference = d

We know :-

$\bf a_n = a+(n-1)d$

$\bullet \ \sf 1^{st} \ term + 3^{rd} \ term = 12$

$\longrightarrow \sf a+a+ (3-1)d = 12 \\\\\longrightarrow 2a+2d = 12 \\\\\longrightarrow a+d = 6 \\\\\longrightarrow d = 6 -a \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...................(1)$

$\bullet \ \sf 1^{st} \ term \times 2^{nd} \ term = 24$

$\longrightarrow \sf a \times [ \ a+ (2-1)d \ ] = 24 \\\\\longrightarrow a\times ( a+d ) = 24 \\\\\longrightarrow a^2+ad = 24 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...................(2)$

Substitute the value of d in eq (2) :-

$\longrightarrow \sf a^2 + a(6-a) = 24 \\\\\longrightarrow a^2+6a -a^2 = 24 \\\\\longrightarrow 6a = 24 \\\\\longrightarrow \bf a = 4$

Substitute a = 4 in eq (1) :-

$\longrightarrow \sf d = 6 - 4 \\\\\longrightarrow \bf d = 2$

$\bullet \sf \ a_1 = 4 \\\\\bullet \ a_2 = 4+2 = 6 \\\\\bullet \ a_3 = 6+2 = 8$

The three terms of AP are 4 , 6 and 8 .