Question

sum of any three consecutive terms of AP is 21 and their product is 231. find three terms of the AP

Answer 1

let no. b a-d,a,a+d
a2q,
a-d+a+a+d=21
3a=21
a=7
now
(a-d)(a+d)a=231
(a2-d2)a=231
49*7-7d2=231
now u can solve for d

Answer 2

Hii friend,

Let the three consecutive terms of AP are (a-d), a , (a+d) .

Therefore,

(a-d) + a + (a+d) = 21

3a = 21

a = 21/3

a = 7

And,

(a-d) × a × (a+d) = 231

a(a²-d²) = 231

7[ (7)² - (d)² ] = 231

7(49-d²) = 231

343 -7d² = 231

-7d² = 231-343

-7d² = -112

d²= -112/-7

d² = 16

d=✓16 = 4 .

Therefore,

a = 7 and d =4

Hence,

Three consecutive terms of AP = (a-d) , a, (a+d) = (7-4) , 7 , (7+4) = 3 , 7 , 11.

HOPE IT WILL HELP YOU..... :-)