the sum of 4th and 8th term of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of AP
Answers
Step-by-step explanation:
tn=a+(n-1)d
a/q
sum of 4th and 8th = a+(n-1)d+a+(n-1)d
24= a+(4-1)d+a+(8-1)d
24= a+3d+a+7d
24= 2a+10d
24= 2(a+5d)
12= a+5d
sum of 6th and 10th = a+(n-1)d+a+(n-1)d
44 = a+(6-1)d+a+(10-1)d
44 = a+5d+a+9d
44 = 2a+14d
44 = 2(a+7d)
22 = a+7d
on subtracting eq.1 form eq.2
22-12=a+7d-(a+5d)
10= a+7d-a-5d
10= 2d
5= d
on putting value of d=5 in eq.1 we get
12= a+5d
12= a+5×5
12= a+25
-13= a
now,
a=-13
d= 5
first term is a=-13
second term is a+d= -8
third term is a+2d= -3
.............666.......
Step-by-step 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!