9th term of an arithmatic progressionvis 9
the sum of 4th and th term and 7th term
is 24 , find the A.p
Answers
= a + (n-1)d
where,
a = the first term of the A.P.
n = number of terms of the A.P.
d = difference of consecutive terms of the A.P.
⇒ = a + (9-1) d
⇒ 19 = a + 8d
⇒ 19 - 8d = a ….i.)
⇒ = a + (4-1) d
⇒ = a + 3d
[Substituting the value of a from eq. i.)]
⇒ = 19 - 8d + 3d
⇒ = 19 - 5d
⇒ = a + (7-1) d
⇒ = a + 6d
[Substituting the value of a from eq. i.)]
⇒ = 19 - 8d + 6d
⇒ = 19 - 2d
Sum of 4th ( ) and 7th term ( ) = 24
⇒ + = 24
[Substituting the value of the terms obtained above]
⇒ (19 - 5d) + (19 - 2d) = 24
⇒ 19 - 5d + 19 - 2d = 24
⇒ 38 - 7d = 24
⇒ - 7d = 24 -38
⇒ - 7d = -14
⇒ d = -14/-7
⇒ d = 2
⇒ a = 19 - 8d
⇒ a = 19 - 8 x 2
⇒ a = 19 - 16
⇒ a = 3
First term = a = 3
Second term = a + d = 3 + 2 = 5
Third term = a + 2d = 3 + 2 x 2 = 3 + 4 = 7
Fourth term = a + 3d = 3 + 3 x 2 = 3 + 6 = 9
Fifth term = a + 4d = 3 + 4 x 2 = 3 + 8 = 11
Sixth term = a + 5d = 3 + 5 x 2 = 3 + 10 = 13
Seventh term = a + 6d = 3 + 6 x 2 = 3 + 12 = 15
The A.P. is -
3, 5, 7, 9, 11, 13, 15.