Question

The sum of first six terms of an arithmetic progression is 42. The ratio of its 10th term to its 30 term is 1 : 3. Calculate the first and the thirteenth term of the A.P.

Answer 1

a=2
d=2

13th term of AP= a+(n-1)d
=a+(13-1)d
=2+12×2
=26

Answer 2

Answer:

$\textb {first term (a) = 6}$

$\textb {Thirteenth term a_{13} = 78}$

Explanation:

Let a and d are first term and common difference of an A.P.

According to the problem given,

i) Sum of the first, six term of A.P = 42
$\boxed {a_{n}=a+(n-1)d}$
=> a + a+5d = 42
=> 2a+5d = 42 ----(1)

ii) The ratio of 10th to its 30th term = 1:3
$\frac{a_{10}}{a_{30}}=\frac{1}{3}$
$\implies\frac{a+9d}{a+29d}}=\frac{1}{3}$
$\implies3\left ({a+9d}\right)=a+29d$
=> 3a+27d = a+29d
=> 3a-a=29d-27d
=> 2a=2d
=> a = d ----(2)
Substitute a=d in equation (1),we get
2d+5d=42
=> 7d = 42
=> d = 42/7
=> d = 6 ----(3)
Therefore,
put d = 6 in equation (2), we get
a = d = 6

ii) Thirteenth term in A.P =
$a_{13}=a+12d$
= 6+12×6
=6+72
= 78

Therefore,
$a_{13}=78$

••••