Question

Find the A.P whose $7^{th}$ and ${13}^{th}$ terms are respectively 34 and 64.

Answer 1

The $n^{th}$ term of AP : $t_n=a+(n-1)d$

According to question, $t_7=34\Rightarrow a+(7-1)d=34\Rightarrow a+6d=\mathrm{34....}\left(1\right)$ and $t_{13}=64\Rightarrow a+(13-1)d=34\Rightarrow a+12d=\mathrm{64....}\left(2\right)$

Subtracting equation 1 from 2, $a-a+12d-6d=64-34\Rightarrow 6d=30\Rightarrow d=30÷6=5$

Now putting the value of d in equation 1, $a+6\times 5=34\Rightarrow a=34-30=4$

Thus, the required AP: $a,a+d,a+2d,a+3d....=4,4+5,4+2\times 5,4+3\times \mathrm{5....}=4,9,14,\mathrm{19...}$