Question

the sum
The first term of
an AP is 2 and
its last ferm is / 158 Then
of its 40th term will be
ST​

Answer 1

Step-by-step explanation:

Given (a) first term of AP = 2

n total terms = 20

(a

n

) least term = 59

We know a

n

=a+(n−1)d

59−2=19d

d=

19

57

=3

∴ common difference of AP is .

For finding 6

th

term from last.

out first term = 52 and last term = 2 & common difference = -3

now

a

6

=a+5d=59+5(−3)

a

6

=59−15

[a

6

=44]

∴6

th

term from last is 44.

Answer 2

Answer:

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