Question

In an arithmetic progression, the sum of first term, third term and the fifth term is $39$ and the sum of second term, fourth term and the sixth term is $51$. Find the tenth term of the sequence

Answer 1

Given : $t_1+t_3+t_5=39$ and $t_2+t_4+t_6=51$

Since, $t_n=a+(n-1)d$

$⟹$ $a+(a+2d)+(a+4d)=39$ and $(a+d)+(a+3d)+(a+5d)=51$

$⟹$ $3a+6d=39$ and $3a+9d=51$

$⟹$ $a+2d=13$ ...... $\left(1\right)$ and $a+3d=17$ ........ $\left(2\right)$

Solving both the equations simultaneously we get,

$d=4$

Substitute the value of $d$ in equation we get,

$a=5$

Now, by the formula $t_n=a+(n-1)d$

$t_{10}=5+(10-1)4=5+9\times 4=41$

$\therefore$ Tenth term of the sequence is $41$.