Question

Find the 4numbers in ap such that the sum of 2nd and 3rd term is 22 and product of 1at and 4the term is 85

Answer 1

Solution :-

Let the four terms of the AP be a, a + d, a + 2d and a + 3d.

We know :-

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

According to the first condition :-

$\longrightarrow \sf (a+d)+(a+2d) = 22 \\\\\longrightarrow 2a+3d = 22 \\\\\longrightarrow 3d = 22-2a \\\\\longrightarrow d = \dfrac{22-2a}{3} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......................(1)$

According to the second condition :-

$\longrightarrow \sf a(a+3d) = 85 \\\\\longrightarrow a^2+3ad = 85 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......................(2)$

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

$\longrightarrow \sf a^2 + 3 \times a \times \dfrac{22-2a}{3} = 85 \\\\\longrightarrow a^2+a(22-2a) = 85 \\\\\longrightarrow a^2+22a-2a^2 = 85 \\\\\longrightarrow -a^2+22a = 85 \\\\\longrightarrow -a^2+22a-85 = 0 \\\\\longrightarrow a^2-22a+85 = 0 \\\\\longrightarrow a^2-17a-5a+85 = 0 \\\\\longrightarrow a(a-17)-5(a-17) = 0 \\\\\longrightarrow (a-5)(a-17) = 0 \\\\\longrightarrow \bf a = 5 \ , \ a = 17$

• If a = 5, then $\sf d = \dfrac{22-2 \times 5} {3}$

                                $= \sf \dfrac{22-10}{3}$

                                 $= \sf \dfrac{12}{3}$

                                  $\bf = 4$

     AP = 5 , 9 , 13 , 17

• If a = 17, then $\sf d = \dfrac{22-2 \times 17 }{3}$

                                 $= \sf \dfrac{22-34}{3}$

                                 $=\sf \dfrac{-12}{3}$

                                 $\sf = -4$

     AP = 17 , 13 , 9 , 5