Question

the sum of the 4th and 8th term of an AP
is 24 and the sum of the 6th and 10th terms is 44. find the first three terms of AP

Answer 1

An4+An8=24
a+(4-1)d+a+(8-1)d=24
a+3d+a+7d=24
2a+10d=28 ---------(1)

An6+An10=44
a+(6-1)d+a+(10-1)d=44
a+5d+a+9d=44
2a+14d=44----------(2)

NOW ELIMINATE EQ 1 AND EQ 2
2a+10d=24
2a+14d=44
-4d= -20
d=20/4
d=5 ----------(3)

NOW SUBSTITUTE EQ 1 AND EQ 3
2a+10(5)=24
2a+50=24
2a=24-50
2a=26
a=26/2
a=13 ---------(4)

First term=a=13
Second term=a+d
=13+5
=18
Third term=2a+d
=2(13)+5
=26+5
=31

So the first three terms are 13,18,31

I HOPE IT WILL HELP U

Answer 2

Answer:

Step-by-step explanation:

Let first term be a and common difference be d.So,4th term=a+3d8th term=a+7dHence,2a+10d=24a+5d=12---------(1)6th term=a+5d10th term=a+9d2a+14d=44a+7d=22--------(2)Solve the two equations, you will get d=5,a=-13First 3 terms=a,a+d,a+2dSubstitute the values of a and d, you will get..First term= -13Second term=-13+5= -8Third term= -8+5= -3Hope you understood,if not ask me