Question

the sum of first six term of an ap is 42 the ratio of its 10 term of its 30 term is 1 ratio 3 calculate the 1st and 13 term of an ap

Answer 1

Let a1  = the first term....we have....the sum of the 1st 6 terms  = 42....therefore...

42 = (6/2)[2a1 + 5d]

42 = 3[2a1 + 5d]

14 = 2a1 + 5d  →   a1 = [14 - 5d]/2 

And a30 = 3(a10)   .... therefore.....

3(a10) = a10 + 20d

2(a10) = 20d

a10 = 10d

So we have

a10 = (a1 + 9d)  ... and by substitution.....

10d = ([14- 5d]/2 + 9d)

20d = 14 - 5d + 18d

20d = 14 + 13d

7d = 14

d = 2

Therefore....

a1 = [14 - 5(2)] / 2  = [14 - 10] / 2 =  4/2 = 2

And

a13 = [2 + 2(12)] =  26

            hope it will help so plzz mark as brainleist one ^_^