the sum of the 4th and 8th term of an ap is 24 and the sum of the 6th and 10th term is 44 find the first three term of the AP
Answers
Answer:
-13, -8, -3
Explanation:
a4 + a8 = 24
a6 + a10 = 44
we know that, in AP, an= a + (n-1)d
a4 + a8 = 24
a + (4-1)d + a + (8-1)d =24
2a + 3d+ 7d = 24
2a = 24 - 10d
2a = 2(12 - 5d)
a = 12 - 5d
a6 + a10 = 44
a + (6-1)d + a+ (10-1)d = 44
2a + 5d + 9d = 44
2a +14d =44
2a = 44 - 14d
2a = 2(22 - 7d)
a = 22 -7d
from the simplified equations....
12 - 5d = 22 - 7d
7d - 5d = 22 - 12
2d = 10
d = 10/2
d = 5
putting the value of 'd' in any of the equations...
a = 12 - 5d
a = 12 - 5(5)
a = 12 - 25
a = -13
a1= -13
a2= a + (n-1)d
a2= -13 +(2-1)5
a2= -13 +5
a2= -8
a3=a+(n-1)d
a3= -13 +(3-1)5
a3= -13+10
a3= -3
Hence, the first three terms of the AP is...
-13, -8. -3
Hope it helps!
:)
Explanation:
We know that nth term of an AP an = a + (n - 1) * d
Given that sum of 4th term and 8th term of an AP is 24.
(i) 4th term = a + (4 - 1) * d = a + 3d.
(ii) 8th term = a + (8 - 1) * d = a + 7d.
Now,
⇒ a + 3d + a + 7d = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 -------- (1)
Given that 6th term and 10th term of an AP is 44.
(i) 6th term a6 = a + (6 - 1) * d = a + 5d.
(ii) 10th term a10 = a + (10 - 1) * d = a + 9d
Now,
⇒ a + 5d + a + 9d = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22 -------- (2)
On solving (1) & (2), we get
⇒ a + 5d = 12
⇒ a + 7d = 22
-----------------
2d = 10
d = 5.
Substitute d = 5 in (1), we get
⇒ a + 5d = 12
⇒ a + 5(5) = 12
⇒ a + 25 = 12
⇒ a = -13.
Hence,
First term = -13.
Second term = -13 + 5 = -8.
Third term = -8 + 5 = -3.
Therefore, the AP is -13,-8,-3.
Hope this helps!