Question

in an AP T10= 20. T20=10. find T30
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Answer 1

ANSWER :

T30= 0

KINDLY CHECK THE ATTACHMENT FOR THE STEPWISE SOLUTION AND ANSWER.

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More Formulae :-

1. Tn = a +(n-1)d

2. Sn = n/2 [2a+(n-1)d]

3. Sn = n/2 (a+l)

4. 2b = a+c

where => Tn = nth term

n = number of terms

a = first term

d= common difference

l = last term

b = second term

Answer 2

Answer:

0

Step-by-step explanation:

Tn=a+(n-1)d

T10=a+(10-1)d T20=a+(20-1)d

T10=a+9d T20=a+19d

20=a+9d 10=a+19d

20-9d=a.......(1) 10-19d=a .......(2)

from (1) and (2)

20-9d=10-19d

19d-9d=10-20

10d=(-10)

d=(-1)

putting d=(-1) in equation (1)

20-9d=a

20-9(-1)=a

20+9=a

a=29

putting a and d in equation Tn=a+(n-1)d

T30=29+(30-1)(-1)

=29+29(-1)

=29-29

=0

T30=0