Question

the 39th term of arithmetic sequence if a1=40 and d=1_2​

Answer 1

$Given \: first \:term (a) = 40 , \:and \\common \: difference (d) = 12 \: in \: an \:A.P$

$\boxed{\pink{ n^{th} \:term (a_{n}) = a+(n-1)d}}$

$\red{ 39^{th} \:term } \\= a_{39} \\= 40 + (39-1)\times 12 \\= 40 + 38 \times 12 \\= 40 + 456\\= 496$

Therefore.,

$\red{ 39^{th} \:term }\green{ = 496}$

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Answer 2

Answer:

Givenfirstterm(a)=40,and

commondifference(d)=12inanA.P

\boxed{\pink{ n^{th} \:term (a_{n}) = a+(n-1)d}}

n

th

term(a

n

)=a+(n−1)d

\begin{gathered} \red{ 39^{th} \:term } \\= a_{39} \\= 40 + (39-1)\times 12 \\= 40 + 38 \times 12 \\= 40 + 456\\= 496 \end{gathered}

39

th

term

=a

39

=40+(39−1)×12

=40+38×12

=40+456

=496

Therefore.,

\red{ 39^{th} \:term }\green{ = 496}39

th

term=496