# which term of an AP 3,15,27,39............ will 132 more than its 54th term

a = 3

d = 15 - 3 = 12

an = a + (n-1)d

a54 + 132  = 32 + 53 * 12 + 132

= 32 + 636 + 132  = 800

an = 800

800 = 32 + (n-1) 12

768 = (n-1)12

n-1 =738 / 12 = 64

n = 64 + 1 = 65

therefore , 65th term is 132 more than 54th term

• 25

a=3

d=15-3=12

now we know the formula for nth term of an AP=a+(n-1)d

therefore 54th term=a+(n-1)d

=3+(54-1)12=639

now the value of nth term which is 132  more than 54th term =639+132=771

now the required  term is

a+(n-1)d=771

=3+(n-1)12=771

n-1=768/12

n=64+1

n=65

threfore the answer is 64th term

• 49

here

a=3

d=12

so

54th term=3+53*12=639

132+54th term=771

771=3+(n-1)12

n-1=64

n=65

so req.

• 10
a=3
d=a​2-a1=15-3=12
now,   An=a+(n-1)d
a54=3+(54-1)12
a54=3+53*12
a54=3+636
a54=639

now,
a54+132=639+132=771
now apply,

an=a+(n-1)d
771=3+(n-1)*12
771=3+12n-12
771=12n-9
​771+9=12n
780/12=n
n=65
a65=a54+132
hece proved

• 15
First term = 3 and Common difference = 12.
Let nth term of the A.P. be 132 more than its 54th term.
an = 132 + a54
a+ ( n - 1 )d  = 132 + ( a + 53d )
3 + 12 ( n - 1 )d  =132 + ( 3 + 53*12 )
12n - 9 = 771
12n = 780
n = 780
12
n = 65.
Hence, 65th term of the given A.P is 132 more than its 54th term.
• 11
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Now we know the formula for nth term of an AP= a+(n-1)d
Therefore 54th term =3+(54-1)12=639
Now the value of nth term which is 132 more than 54th term = 639+132=771
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therefore the answer is 64th term.
• -1
65th term

• 2
a+[n-1]d = 771
3+[n-1]12=771
3+12n-12=771
12n-12=771-3=768
12n=768+12=780
n=780/12=65
hence, the 65th term is the answer
• -3
64th term
• -3
Hehe PLZZ thumbs up

• 3 • 7
Ans is 65 the term • 2
Ans is 65 the term • 1
Let the required term be nth term 54th term = a +(n-1)d = 3 + 53x12 = 3 + 636 = 639 therefore 132 + 639 = 771 will be the nth term. 771 = 3 + (n-1)12 768/12 +1 = n therefore n = 64 +1 = 65 therefore the 65th term will be 132 more than the 54th term.... ANSWER
• 1
a = 3 d = 15 - 3 = 12 an = a + (n-1)d a54 + 132 = 32 + 53 * 12 + 132 = 32 + 636 + 132 = 800 an = 800 800 = 32 + (n-1) 12 768 = (n-1)12 n-1 =738 / 12 = 64 n = 64 + 1 = 65 therefore , 65th term is 132 more than 54th term
• 1
65th term

• -1
n=65
• 0
Given, a=3 d=12 a54=a+53d =639 A/Q an=a54+132 a+n-1*d=639+132 3+n-1*d=771 n=64+1 n=65 N
• 0
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