Question

plzz solve this...

Which term of the AP ; 3,15,27, 39,... Will be 132 more than its 54th term ?

Answer 1

hiii. ......friend

the answer is here,

=> 3,15,27,39,........

a=3.

d=15-3

=> 12.

54th term,

=> an= a+(n-1)d.

=> 3+(54-1)(12)

=> 3+53 (12)

=> 3+ 636

=>639.

The term is 132 more than 54th term.

=> 132+639.

=> 771 .

So,

=> 771 = a+(n-1)d

=> 771 =3+(n-1)12.

=> 768= 12 (n-1).

=> n-1 =768/12

=> n-1 =64.

=> n=65.

So, The number is 65th term in the list of numbers.


:-)Hope it helps u.

Answer 2

3,15,27,39.......
let find out 54th term

first term = a = 3
common difference = d = T2-T1 = 15-3 = 12
number of terms = n = 54
T54 = ?

Tn = a+(n-1)d
T54 = 3+(54-1)12
T54 = 3+(53)12
T54 = 3+636
T54 = 639

let take that term as Tn then

Tn = T54+132
Tn = 639+132
Tn = 771

let find 771 is which term of A.P

Let take
Tn = 771
a = 3
d = 12
n = ?

Tn = a+(n-1)d
771 = 3+(n-1)12
771-3 = 12n-12
768 = 12n-12
768+12 = 12n
780 = 12n
780/12 = n
65 = n

so " 65th " term of (A.P 3,15...... )will be 132 more than its 54th term