Question

Find the 12th term of each sequence. a9= 120, a14=195

Answer 1

The general formula for an Arithmetic Progression is :

                            $\boxed {\boxed {T_n = a_1 + (n-1)d}}$


$\text {Given that } a_9 = 120,$

$T_9 = a_1 + (9 - 1)d = 120$

$a_1 + 8d = 120 \text { ------------------(1)}$


$\text {Given that } a_{14} = 195,$

$T_9 = a_1 + (14 - 1)d = 195$

$a_1 + 13d = 195 \text { ------------------(2)}$


Equation (2) - Equation (1) :

$13d - 8d = 195 - 120$

$5d = 75$

$d = 15$


Substitute d = 15 into Equation (1) :

$a_1 + 8(15) = 120$

$a_1 = 120 - 120$

$a_1 = 0$

$\text {Substitute }a_1 = 0 \text{ and d = } 15,$

               $\boxed { \boxed { T_n = 15(n -1) }}$


Find 12th term :

$T_{12} = 15(12 -1) = 165$


$\boxed {\boxed {\text {Answer: The 12th term is 165.}}}$