24. The sum of first six terms of an.A.P. is 42. The ratio of its 10th

1.	24. The sum of first six terms of an.A.P. is 42. The ratio of its 10th30th.term is 1 :3. Calculate the first and 13th term of the A.P
Answer» 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

24. The sum of first six terms of an.A.P. is 42. The ratio of its 10th30th.term is 1 :3. Calculate the first and 13th term of the A.P

Answer»

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