Question

The 5th term of an A.P is 1 whereas its 31st term is - 77. Then its 20th term is ______ .

Answer 1

Answer:

Given T

5

=a+4d=1,T

31

=a+30d=−77,

Solving the above

two, we get a=13 and d=−3.

T

20

=13−(20−1)3=−44

S

15

=[n/2][2a+[n−1]d]=[15/2][26+14[−3]]=−120

Let T

n

=−17.

Then a+[n−1]d=−17. 13+[n−1][−3]=17∴3n=33

or n=11

Let S

n

=20 Then (n/2)[2a+(n−1)d]=20.

n[n×13+(n−1)(−3)=40] by [1].

3n

2

−29n+40=0∴[n−8][3n−5]=0∴n=8

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Answer 2

Answer:

5th term = 1

=> a + 4 d = 1               =>(1)

31st term = - 77

=> a + 30 d = - 77        =>(2)

By substacting eq 2 by eq 1 , we get,

  a + 30 d = - 77

  a + 4   d =     1

_____________

        26 d = - 78

=> d = -78 / 26

=> d = - 3

sub d = - 3 in eq(1), we get,

a + 4 (-3) = 1

=> a = 13

So 20th term = a + 19 d = 13 + 19 * (-3)

                                         = - 44

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