Question

what is value of t25 and t5 for A.P 9,14,19.... give explinetion
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Answer 1

Answer:

The ap is

9,14,19...

t1=9

d=14-9=5

t25=a+(25-1)×d

=9+(24×5)

=9+120

=129

t5=a+(5-1)×d

=9+(4×5)

= 9+20

=29

note that 'n'th term =a+(n-1)×d

Step-by-step explanation:

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Answer 2

${\large{\text{\underline{\underline{\red{Given :}}}}}}$

$\text{9,14,19 are in A.P}$

${\large{\text{\underline{\underline{\red{To Find :}}}}}}$

$t_{25}\:\:and\:\:t_5$

${\large{\text{\underline{\underline{\red{Formula Applied :}}}}}}$

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

${\large{\text{\underline{\underline{\red{Solution :}}}}}}$

$\text{It's given that 9,14,19 are in A.P hence,}$

$a \implies \text{first term} \implies \bf 9$

$d\implies\text{common difference} \implies t_2-t_1 = 14-9 \implies \bf 5$

$\bigstar{\text{\red{Let's Find t_5 }}}$

$n\implies \text{No.of.term} \implies 5$

$\implies t_n = a+(n-1)d$

$\implies t_5=9+(5-1)5$

$\implies t_5 = 9+(4)5$

$\implies t_5=9+20$

$\implies t_5 = 29$

$\bigstar{\red{\text{Let's find t_{25} }}}$

$\implies t_{25}=a+(n-1)d$

$\implies t_{25}=9+(25-1)5$

$\implies t_{25}=9+(24)5$

$\implies t_{25} = 9+120$

$\implies t_{25} = 129$