Question

(3) Find four consecutive terms in an A. P. such that
their sum is 58 and the product of the second and the
third term is 208.​

Answer 1

let the 4 terms

= a-3d , a - d, a +d , a+3d

sum= a + 3d + a+d + a-3d +a -d

58 = 4a

a= 14.5

product of second and third = 208

(a+d )(a-d) = 208

a²- d²= 208

(14.5)² -208 = d²

210.25 -208 = d²

d= ✓ 2 . 25

=± 1.5

thus, the nos , when d= + 1.5

14.5 - 3( 1.5)= 10

14.5 - ( 1.5)= 13

14.5 + ( 1.5) = 16

14.5 +3( 1.5) = 19

when d = - 1.5,

the nos get reversed..

19 , 16, 13, 10