Question

for the ginite AP 5,9,13,...101, find the 9th term from the end.

Answer 1

Heya mate, Here is ur answer

a=5

d=9-5

d=4

an=l=101

101=a+(n-1)d

101=5+(n-1)×4

101-5=(n-1)×4

96=(n-1)×4

96/4=n-1

24=n-1

24+1=n

25=n

For finding 9th term from last , An will become first term

l9=l-(9-1)d

=101-8×4

=101-32

=69

So,9th term from the last is 69 .

Warm regards

@Laughterqueen

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Answer 2

Let:

a be the first term

d be the common difference

n be the number of terms

Given:

a=5

d=9-5=4

We know that nth term of an AP is a+(n-1)d

==)101=a+(n-1)d

==)5+4(n-1)=101

==)4(n-1)=101-5

==)4(n-1)=96

==)n-1=96/4

==)n-1=24

==)n=24+1

==)n=25

Lets do a trick.

Just reverse the AP.

101,97,95....................5

It has d as -4.

9th term from the end of that AP will be the 9th term of this AP

So:

a+(n-1)d

==)101+(9-1)(-4)

==)101-8*4

==)101-32

==)69

Okay 9th term is an interesting number 69.

Hope it helps.