Question

Find the term of the arithmetic progression 9, 12, 15, 18, ... which is 39 more than its 36th term.

Answer 1

Answer:

49th term of the A.P   9, 12, 15, 18, ... which is 39 more than its 36th term.

Step-by-step explanation:

Given :  

A.P  9, 12, 15, 18, ..

first term , a = 9 , common difference , d = 12 - 9 = 3

nth term = 39 + a36

a + (n -1)d = 39 + a36

[nth term = a + (n -1)d]

 9 + (n - 1)3 = 39 + a + (36 - 1) d

9 + 3n - 3 = 39 + 9 + 35 ×  3

6 + 3n  = 39 + 9 + 105

6 + 3n = 48 +105

6 + 3n   = 153

3n = 153 - 6

3n = 147

n = 147/3

n = 49

Hence, 49th term of the A.P   9, 12, 15, 18, ... which is 39 more than its 36th term.

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Answer 2

Step-by-step explanation:

1st term =>a =9.

Common difference=>(12-9)=>3=d.

nth term=>a+(n-1)d

36th term =>9+(35)3=>114.

150 is the required number.150=9+(n-1)3=>

150=6+3n=>144=3n=>n=144/3=48. So the 48th term of the Arithematic progression 9,12,15,18,21,24,27,30,33,36,39,.... is 150.

Hope it helps you.