find the fifthteen term of ap a=-23 d=4
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Answer:
The nth term or general term of an A.P is used to find terms arithmetic progression.
Let us consider an A.P. with first term as 'a' and the common difference as 'd'.
So the sequence is a, a + d, a + 2d, a + 3d, ... then,
1st term = a1 = a + (1 - 1)d
2nd term = a2 = a + d = a + (2 -1)d
3rd term = a3 = a + 2d = a + (3 -1)d
4th term = a4 = a + 3d = a + (4 -1)d
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nth term = an = a + (n- 1)d
From the above we observe that the terms of A.P is given by
First term + ( term number - 1 ) (common difference)
Formula to find nth term of an A.P sequence is
an = a + ( n - 1) d
For example 15th term will be :
a15 = a + (15 - 1)d = a + 14d
Examples on Find terms arithmetic progression
1) Find the 19th and 24th terms of the A.P given by 21,16,11,6,1,...
Solution : We know that nth term of an A.P. is given by
an = a + (n - 1) d
To find a19,
n = 19, a = 21 and d = 16 - 21 = - 5
a19 = 21 + ( 19 -1 )(-5)
= 21 + 18 x(-5)
= 21 - 90
∴ a24 = - 69
To find a24,
n = 24, a = 21 and d = 16 - 21 = - 5
a24 = 21 + ( 24 -1 )(-5)
= 21 + 23 x(-5)
= 21 - 115
∴ a24 = - 94
2) Which term of the A.P. is 1, 6,11,16,... is 301 ?
Solution : The given sequence is in A.P.
So, a = 1 and d = 6 -1 = 5
Let us consider nth term as 301
an = a + (n - 1) d
301 = 1 + (n - 1)(5)
301 = 1 + 5n - 5
301 = -4 + 5n
301 + 4 = 5n
305 = 5n
∴
305
5
=
5n
5
∴ n = 61
Thus 301 is 61st term.
3) Which term of the A.P. 5,15,25,... will be 130 more than its 31st term.
Solution: We have, a = 5 and d = 15 - 5 = 10
an = a + (n - 1)d
so, 31st term is
a31 = 5 + (31 - 1)(10)
a31 = 5 + 30 x 10
a31 = 305
Let the nth term of the given A.P. be 130 more than its 31st term.
an = 130 + a31
an = 130 + 305
an = 435
5 + (n - 1)(10) = 435
5 + 10n - 10 = 435
- 5 + 10n = 430
10n = 440
10n
10
=
440
10
∴ n = 44
Hence, 44th term of the given A.P is 130 more than its 31st term.