Question

the sum of first 6 terms of an ap is 42 the ratio of 10th term to the 30th term is 1 is to 3 find the first term and the 11th term of an ap

Answer 1

Hey

Here is your answer,

Tn = [a + (n-1)d]
Sn = n/2*[2a +(n-1)d].
Given
T30/T10 = 3
(a + 29d)/(a + 9d) = 3
After simplification we get
a = d

S6 = 6/2 [2a + 5d] = 42 (given)
14 = 2a + 5d
14/7 = 2 = a = d ( substituting a = d)

So the series starts like
2, 4, 6, 8, 10, 12, 14, 16, .......
Now T11 = [2+10d] = [2 + 20 ] = 22.

Hope it helps you!

Answer 2

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

I HOPE THIS HELPS YOU!