Question

which term of the AP 3 15 27 39.... will be 120 more than its 21st term

Answer 1

First term = 3

Common difference :-

=> $t_{2}$ - $t_{1}$

= 15 - 3

= 12

Hence :-

21st term = a + 20d

= 3 + 240

= 243

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

=> 243 + 120 = 3 +(n - 1) × 12

=> n - 1 = $\frac{360}{12}$

=> n - 1 = 30

=> n = 30 + 1

=> n = 31

Answer 2

Answer:

$31^{st}\ term$

Step-by-step explanation:

$Let\ the\ term\ be\ n^{th}\ term\ of\ the\ AP. \\ \\ \\ d = 15 - 3 = 12 \\ \\ \\ T_n - T_{21} = 120 \\ \\ = (n - 21)d = 120 \\ \\ = (n - 21)12 = 120 \\ \\ n - 21 = \frac{120}{12} = 10 \\ \\ n = 10 + 21 = \bold{31}$

$\\ \\ \\ \therefore\ \bold{31}^{st}\ term\ of\ the\ AP\ 3, 15, 27, 39,...\ is\ 120\ more\ than\ its\ 21^{st}\ term. \\ \\ \\ Hope\ this\ may\ be\ helpful. \\ \\ Please\ mark\ my\ answer\ as\ the\ \bold{brainliest}\ if\ this\ may\ be\ helpful. \\ \\ Thank\ you.\ Have\ a\ nice\ day. \\ \\ \\ \#adithyasajeevan$