If the third term of an A.P is 7
and the 6th term is 18 then find
the sum of first 5 terms.
Answers
ATQ:
3rd term of an AP = 7
We know that the third term can be written as:
➞ a₃ = 7
We know that a = a + (n - 1)d.
➞ a + (n - 1)d = 7
➞ a + (3 - 1)d = 7
➞ a + 2d = 7 ⇒ Eq(1)
It has also been given that the 6th term is 18.
We know that the sixth term can be written as:
➞ a₆ = 18
We know that a = a + (n - 1)d.
➞ a + (n - 1)d = 18
➞ a + (6 - 1)d = 18
➞ a + 5d = 18 ⇒ Eq(2)
Subtracting Eq(1) from Eq(2) we get:
➞ a + 5d - (a + 2d) = 18 - 7
➞ a + 5d - a - 2d = 11
➞ 5d - 2d = 11
➞ 3d = 11
➞ d = 11/3
Substitute the value of 'd' in Eq(1).
➞ a + 2d = 7
➞ a + 2(11/3) = 7
➞ a + 22/3 = 7
➞ a = 7 - (22/3)
➞ a = (21 - 22/3)
➞ a = -1/3
Now, Let's find the sum of the first 5 terms.
Here, n = 5.
Therefore, the sum of the first 5 terms is 35.
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Definition of the terms used:
a ➞ First term of the Arithmetic Progression.
d ➞ Common difference of the Arithmetic Progression.
a ➞ Position of the nth term.
S ➞ Sum of the first n terms.
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