Question

the sixth term of an a.p. is 12 and its eighth term is 22. locate its first term, normal contrast and sixteenth term.

Answer 1

$<b><u><i>$Let, first term = a and normal contrast =d.

T6 = 12 ⇒ a+5d= 12 …. (i)


T8= 22 ⇒ a+7d = 22 … (ii)


On subtracting (i) from (ii)


we get 2d = 10


⇒ d = 5


Putting d= 5 in (i), we get a+5×5 = 12


⇒ a= (12-25) = - 13


∴ First term = - 13


▶️normal distinction = 5


▶️T16= a+ 15d = - 13 + 15×5 = (75 -13) = 62


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