Question

if the sum of three consecutive term of an ap is 51 and the product of two extremes is 273 find the number​

Answer 1

Answer:

Let 3 consecutive terms A.P is a –d, a , a + d. and the sum is 51

so, (a –d) + a + (a + d) = 51

> 3a –d + d = 51

> 3a = 51

> a = 17

The product of first and third terms is 273

So  it stand for ( a –d) (a + d) = 273

 >a 2 –d 2 = 273

> 172 –d 2 = 273 

>  289 –d 2 = 273

>  d 2 = 289 –273

>  d 2 = 16

>  d = (plus and minus) 4

so now put plus and minus eventually to get the third term. 

Answer 2

Answer:

A+a+d+a+2d=3a+3d =51, a+d=17 so product of extremes is a(a+2d)=273 so a=17-d (17-d)(a+2d)=(17-d)(17+d)=289-d2=273=d2=16 so d=4 now a=17-4=13 so numbers are 17,21,25