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

Given that,
Second term, $$a_2$$ = 14
Third term, $$a_3$$ = 18
Common difference, d
= $$a_3 - a_2$$ = 18-14 = 4
$$a_2$$ = a + d
14 = a + 4
a = 10 = First term

Sum of n terms;

$$S_n = {{n} \over {2}} [2a + (n – 1)d]$$
$$S_{51} = {{51} \over {2}} [2×10 (51-1) 4]$$
= $${{51} \over {2}}[2 + (20)4$$]
= 51×$${{220} \over {2}}$$
= 51×110
= 5610