If sum of 3 consecutive terms of an ap is 51 and product of 1st and 3rd term is 273
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Let a-d be the first term
a be the second term
a+d be the third term
So the a. p. is a-d,a, a+d
Now,
a-d+a+a+d=51
3a=51
a=51/3
a=17
Now,
(a-d) *(a+d) =273
a^2-d^2=273
17^2-d^2=273
289-d^2=273
-d^2=273-289
-d^2=-16
d^2=16
d=√16
d=4
So
a-d=17-4
a-d=3(1st term)
a=17(2nd term)
a+d=17+4=21
Hence, we got A. P. 3,17,21.
#snj
a be the second term
a+d be the third term
So the a. p. is a-d,a, a+d
Now,
a-d+a+a+d=51
3a=51
a=51/3
a=17
Now,
(a-d) *(a+d) =273
a^2-d^2=273
17^2-d^2=273
289-d^2=273
-d^2=273-289
-d^2=-16
d^2=16
d=√16
d=4
So
a-d=17-4
a-d=3(1st term)
a=17(2nd term)
a+d=17+4=21
Hence, we got A. P. 3,17,21.
#snj
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