Question

the sum of 6th term of an ap is 42 the 10 the term and 30 term are ine the ratio 1:2 find the 13 term of an ap​

Answer 1

Answer:

Sum of the first n terms is given by

Sn=2n[2a+(n−1)d]

S6=26[2a+(6−1)d]

S6=3[2a+5d]

T30T10=31

a+29da+9d=31

(a+9d)3=(a+29d)1

3a+27d=a+29d

2a=2d

a=d...(1)

S6=3[2a+5d]

S6=3[2a+5a]

42=3(7a)

42=21a

∴a=2

From (1)

d=a=2

T13=a+(n−1)d

=2+

Answer 2

Question:

the sum of 6th term of an A.P is 42, 10th term and 30th term are in the ratio 1:2 find the 13th term of an A.P.

Solution:

$T_n = a + (n-1)d \\\\ 6th~term= a+5d = 42 - - - [i]$

And,

$\frac{10th~term}{30th~term} = \frac{a+9d}{a+29d} = \frac{1}{2} \\\\ 2a+18d= a+29d \\\\ 2a-a+18d-29d=0 \\\\ a-11d=0 - - - [ii]$

Subtracting [ii] from [i], we get :

$a + 5d - a +11d =42 \\\\ 16d = 42 \\\\ d= \frac{21}{8} - - - [iii]$

Putting value of d in [i], we get:

$a + 5 \times \frac{21}{8} = 42 \\\\ a + \frac{105}{8} = 42 \\\\ a = 42 - \frac{105}{8} \\\\ a = \frac{336-105}{8} = \frac{231}{8} - - - [iv]$

Now,

$13th~term = a + 12d \\\\ =\frac{231}{8} + 12 \times \frac{21}{8} \\\\ = \frac{231 +252} {8} \\\\ \frac{483}{8}$

There's a mistake in your question i think, the ratio given is 1:2, but it should be 1:3.

Then, answer would have been a natural number instead of fraction.