Question

The sum of six term of an AP is $42$. The ratio of its ${10}^{th}$ term to its ${30}^{th}$ term is $1:3$. Find the first and ${13}^{th}$ term.

Answer 1

$s_6=42\frac{a_{10}}{a_{30}}=\frac{1}{3}\frac{a+9d}{a+29d}=\frac{1}{3}3(a+9d)=a+29d3a+27d=a+29d2a=2da=d.......\left(1\right)s_6=42s_n=\frac{n}{2}[2a+(n-1\left)d\right]=>s_6=3[2a+(6-1\left)d\right]42=3[2a+5d]42=3[2a+5a][\because a=d]42=21aa=2=d$

$\therefore$ first term $=2$

${13}^{th}$ term $a_{13}=a+12d=2+24=26$