Math, asked by manoj10, 1 year ago

find the 10th term from end of the ap 8,10,12...126


shilpak: 126 = 8+(n-1)2
118 = (n-1)2
59 = n-1
n = 60

the pth term from the last

t(n-p+1) = a+(n-p)d
t(60-10+1) = 8 +(60-10)2
t51 = 108

Answers

Answered by mysticd
308
Hi ,

8 , 10 ,12 , ..., 124 , 126 is an A.P

Rearranging the given A.P in

reverse order we get,

126 , 124 , ....., 12 , 10 , 8

First term = a = 126

Common difference = a2 - a1

d = 124 - 126

d = -2

n = 10

____________________________

We know that ,

n th term = an = a + ( n - 1 ) d

____________________________

10th term = a10

a10 = a + 9d

= 126 + 9 ( - 2 )

= 126 - 18

a10 = 108

Therefore ,

10th term from the end of the A.P

= 108

I hope this helps you.

***
Answered by Mohanaanagaraj21
148

all the best for your board exams......

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