Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively

Solution:

$\mathrm{t}_{2}=14, \mathrm{t}_{3}=18$

$\mathrm{d}=\mathrm{t}_{3}-\mathrm{t}_{2}=18-14=4$, i.e., $\mathrm{d}=4$

Now $\quad \mathrm{t}_{2}=14 \quad \Rightarrow \mathrm{a}+\mathrm{d}=14$

$\Rightarrow a+4=14 \quad \Rightarrow a=10$

$\mathrm{S}_{51}=\frac{51}{2}\{2 \mathrm{a}+50 \mathrm{~d}\}=\frac{51}{2}\{2 \times 10+50 \times 4\}$

$=\frac{51}{2} \times 220=51 \times 110=5610$