Question

Find four consecutive terms in an AP whose sum is 12 and the product of 3rd and 4th term is 45​

Answer 1

Answer:

let the terms be a-d,a,a+d,a+2d...

sum=4a+2d=12

2a+d=6...... (eqn1st )....

sum of 3rd and 4th term...

a+d,+(a+2d)=14

2a+3d=14....(eqn 2nd)

by subtracting 1st from 2nd

-2d=-8

d=4

from eqn 1st a=1

so the terms are (1-4),1,(1+4),(1+2 (4)=

-3,1,5,9.......