4th term of an arithmetic sequence is 23 and its 11th term 65. what is its common difference? what is its first term? find the position of 299 in this sequence
Answers
Answer:-
Given:
4th term of an AP = 23
11th term = 65
We have to find first term and common difference.
We know that,
nth term of an AP – a(n) = a + (n - 1)d
Hence,
→ a + (4 - 1)d = 23
→ a + 3d = 23 -- equation (1)
Similarly,
→ a + 10d = 65 -- equation (2)
Subtracting equation (1) from (2) we get,
→ a + 10d - (a + 3d) = 65 - 23
→ a + 10d - a - 3d = 42
→ 7d = 42
→ d = 42/7
→ d = 6
Putting the value of d in equation (1) we get,
→ a + 3 * 6 = 23
→ a = 23 - 18
→ a = 5
Also,
We have to find which term of the AP is 299.
So let a(n) = 299
Hence,
→ 299 = a + (n - 1)d
→ 299 = 5 + (n - 1)(6)
→ 299 - 5 = 6n - 6
→ 299 - 5 + 6 = 6n
→ 300 = 6n
→ 300/6 = n
→ 50 = n
Therefore,
- first term = 5
- common difference = 6
- 299 is the 50th term of given AP.
Answer:
comman difference (d) = 6
first term (a) = 5
position of 299 = 50
Step-by-step explanation:
Given :
4th term = 23
i.e. a+3d = 23 _____equation [1]
11th term = 65
i.e. a+10d =65 ______equation [2]
Now subtracting the equations
a + 3d - (a+10d) = 23 - 65
a + 3d - a - 10d = - 42
-7d = - 42
d = 6
To get the value of a i.e first term, putting the value of d in equation 2.
i.e. a + 10(6) = 65
a = 65 -60
a = 5
Position of 299
299 = a + ( n-1 ) d
299 = 5 + (n-1) 6
294/6 = n -1
49 + 1 = n
50 = n
299 is in the 50th position
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