Find the term of an ap is 9,12,15,18....... Which is 39 more than its 36th term
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Solution :-
Given A.P. = 9, 12, 15, 18...........
a = 9
d = 12 - 9
d = 3
Now, according to question.
a₃₆ = a + (n - 1)d
a₃₆ = 9 + (36 - 1)3
= 9 + 35*3
= 9 + 105
a₃₆ = 114
The required term is 39 more than its 36th term.
So,
⇒ a₃₆ + 39
⇒ 114 + 39 = 153
an = 153
153 = a + (n - 1)d
153 = 9 + (n - 1)3
153 = 9 + 3n - 3
3n = 153 + 3 - 9
3n = 147
n = 147/3
n = 49
So, the required term is 49th term of the given A.P.
Answer.
Given A.P. = 9, 12, 15, 18...........
a = 9
d = 12 - 9
d = 3
Now, according to question.
a₃₆ = a + (n - 1)d
a₃₆ = 9 + (36 - 1)3
= 9 + 35*3
= 9 + 105
a₃₆ = 114
The required term is 39 more than its 36th term.
So,
⇒ a₃₆ + 39
⇒ 114 + 39 = 153
an = 153
153 = a + (n - 1)d
153 = 9 + (n - 1)3
153 = 9 + 3n - 3
3n = 153 + 3 - 9
3n = 147
n = 147/3
n = 49
So, the required term is 49th term of the given A.P.
Answer.
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