Question

Which term of the AP 3, 15, 27, 39, .... will be 144 more than its 54th term ?​

Answer 1

3, 15, 27, 39…….

an=a+(n−1)d

a=3,d=15−3=12

n=54

a54=a+(n−1)d

=3+(54−1)×12

=3+53×12

=639

Term which is 144 more than its 54th term is – 639 + 144 = 783

an=773

783=a+(n−1)d

783=3+(n−1)12

783-3=12n-12

780+12=12n

792=12n

n=792/12

n=66

Answer 2

Answer:

66th term

Step-by-step explanation:

Given:-

$a=3$

$d=a_{2}-a$

$d=15-3$

$d=12$

$x=a_{54}+144$, where $x$ is the $n^{th}$ term which will be 144 more than the 54th term.

To find:-

$n$

Solution:-

$a_{n}=a+(n-1)d$

$a_{54}=a+(54-1)d$ (Here $n=54$)

$a_{54}=3+53(12)$

$a_{54}=3+636$

$a_{54}=639$

$x=a_{54}+144$

$x=639+144$

$x=783$

$a+(n-1)d=x$

$3+(n-1)12=783$

$(n-1)12=783-3$

$(n-1)12=780$

$n-1=\frac{780}{12}$

$n-1=65$

$n=65+1$

$n=66$

∴ The 66th term of the given AP will be 144 more than the 54th term.