Question

The sum of first six terms of an A.P. is 42. The ratio of 10th term to its 30th term is 1:3. Calculate the 1st and 13th term of the A.P.

• A. 1st term=0, 13th term=26
• B. 1st term=1, 13th term=25
• C. 1st term=2, 13th term=26
• D. 1st term=2, 13th term=24

Answer 1

The correct option is C 1st term=2, 13th term=26

Let the first term be 'a' and common difference be 'd'.
$S_6=42$
$\Rightarrow \frac{n}{2}(2a+(n-1\left)d\right)=42$
$\Rightarrow \frac{6}{2}(2a+(6-1\left)d\right)=42$
$\Rightarrow 2a+5d=14$ .............(1)
$\frac{t_{10}}{t_{30}}=\frac{1}{3}$
$\Rightarrow \frac{a+9d}{a+29d}=\frac{1}{3}$
$\Rightarrow 3(a+9d)=1(a+29d)$
$\Rightarrow 3a+27d=a+29d$
$\Rightarrow 3a-a=29d-27d$
$2a=2d$
$a=d$

Substituting a=d in (1), we get:
$2a+5d=14$
$\Rightarrow 2d+5d=14\Rightarrow 7d=14$
$d=2=a$
1st term $=t_1=a=2$
13th term = $t_{13}=a+12d$
$=2+\left(12\right)2$
$t_{13}=26$