Question

The nth term of a sequence is given by an= 2n+7. show that it is an A.P. Also, find its 7th term.

Answer 1

first we take n=1,2,3...
then a1=2*1+7=9
a2=2*2+7=11
a3=2*3+7=13
a2-a1=11-9=2
a3-a2=13-11=2
their common difference is 2 which is same so this is an ap
we have a=9
d=2
so a7=a+(n-1)*d
=9+(7-1)*2
=9+6*2
=9+12
=21

Answer 2

The 7th term of an AP is 21.

Step-by-step explanation:

We have,

$a_{n} =2n+7$

To, find $a_{7} =?$

Put n = 1 , 2 , 3, 4, ...., n

$a_{1} =2(1)+7=2+7=9$,

$a_{2} =2(2)+7=4+7=11$,

$a_{3} =2(3)+7=6+7=13$,

.

.

.

$a_{7} =2(7)+7=14+7=21$

The 7th term of an AP, $a_{7} =2(7)+7=14+7=21$

The given sequence are:

9, 11, 13, ....

Here, first term(a) = 9 and

Common difference$(d) =a_{2} -a_{1}=a_{3} -a_{2}$

= 11 - 9 = 13 - 11 = 2

Hence, the given sequence are in AP.(proved)