Find the term of the arithmetic progression 9,12,15,18.... which is 39 more than its 36th term.
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Answered by
6
Answer:
First term, a = 9, d = 12 – 9 = 3
Let, last term be an
an = a + (n – 1) d
= 9 + (n – 1)3
= 9 + 3n – 3 = 6 + 3n (i)
Let, 36th term, a36 = a + 35d
= 9 + 35 (3) = 114
Now the term is 39 more than its 36th term
an = 39 + a36
= 39 + 114 = 153
Putting the value in (i), we get
153 = 6 + 3N
3n = 153 – 6
3n = 147
n = 49
Hence, 49th term of the given A.P. is 39 more than its 36th term
Answered by
2
Answer:
Answer:
First term, a = 9, d = 12 – 9 = 3
Let, last term be an
an = a + (n – 1) d
= 9 + (n – 1)3
= 9 + 3n – 3 = 6 + 3n (i)
Let, 36th term, a36 = a + 35d
Step-by-step explanation:
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