please answer both the questions.
Answers
hi
Given is that : 3rd term of the AP = 7
7th term = 23
We know that the nth term of an AP = a + ( n - 1 )d
Where,
a = first term of the AP
d = common difference
For the 3rd term, using the formula :
a + ( n - 1 )d = \sf{a_{n}}a
n
The values will be =》 a + ( 3 - 1 )d = 7
=》 a + 2d = 7
=》 a = 7 - 2d
For the 7th term, firstly, we need to take out the value =》 3(7) + 2 = 23
For the 7th term =》 a + ( 7 - 1 )d = 23
=》 a + 6d = 23
=》 a = 23 - 6d
a = a ( the first term will always be the same )
=》 7 - 2d = 23 - 6d
=》 4d = 16
=》 d = 4
a = 7 - 8
a = ( -1 )
Now, we've got the real values of a and d. The 20th term will be :
( -1 ) + ( 20 - 1 )4
=》 ( -1 ) + 76
=》 \boxed{\mathfrak{20th\:term = 75}}
20thterm=75
nth term of the AP = a + ( n - 1 )d
Put the values of a and d :-
=》 ( -1 ) + ( n - 1 )4
=》 4n - 4 - 1
=》 \boxed{\mathfrak{nth\:term=4n - 5}}
nthterm=4n−5
Answer:
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