Which term of an AP $3, 15, 27, 39,$ _______ will be $120$ more than its $21^{s t}$ term.

Open in App

Solution

$\begin{matrix} A . P 3, 5, 27, 39, . . . . . a = 3 [∵ a = f i r s t t e r m] d = 12 [∵ d = c o m m o n d i f f e r e n c e] L e t t e r m b e T_{n} \Rightarrow T_{n} = T_{2} + 120 [g i v e n] \Rightarrow T_{n} = a + (20) d + 120 \Rightarrow T_{n} = 3 + (20) \times 12 + 120 \Rightarrow T_{n} = 3 + (20 + 10) 12 \Rightarrow T_{n} = 3 + (30) 12 \Rightarrow T_{n} = 3 + (31 - 1) 12 T h e n t e r m i s 31 s t t e r m A n s . \end{matrix}$

Suggest Corrections

Similar questions

3, 15, 27, 39, . . .

120

21^{s t}

3, 15, 27, 39, . . . . . . . . .

132

54

3, 15, 27, 39, . . . .

120

21

Join BYJU'S Learning Program