Question

The 25th term of an arithmetic sequence is 140 & 27 th term is 166 write the sequence ​

Answer 1

Answer:

25th term of A.P. is 140 and 27th term is 166

So 26th term will be (140+166) / 2=153

Now common difference(d) is 153–140=13

We know that tn= a+(n-1) d

So a=tn-(n-1) d=140-24*13

=140-312

= –172

Now sequence will be,

–172,–159,–146,–133,...........

If you satisfy on my answer then give me thanks

Answer 2

a₂₅ = 140

a + 24d = 140 ___(i)

Also,

a₂₇ = 166

a + 26d = 166 ___(ii)

Subtracting (ii) equation from (i) equation :

 a + 24d =  140

- a ± 26d = -166

--------------------------

-2d = -26

d = 26/2 = 13

Putting value of d in (i) equation :

a + 24 (13) = 140

a + 312 = 140

a = 140 - 312

a = -172

Sequence :  -172 , -159 , -146 , ......