✒✔✔The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the three terms of the AP. ✔✔✒
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Answers
Answer:
-13, -8, -3
Step-by-step explanation:
Let 'a' be the first term and 'd' be the common difference.
(i) Sum of 4th and 8th terms is 24:
∴ a₄ + a₈
⇒ 24 = [a + (4 - 1) * d] + [a + (8 - 1) * d]
⇒ 24 = [a + 3d] + [a + 7d]
⇒ 24 = 2a + 10d
⇒ a + 5d = 12
(ii) Sum of 6th and 10th terms is 44:
∴ a₆ + a₁₀
⇒ 44 = [a + (6 - 1) * d] + [a + (10 - 1) * d]
⇒ 44 = [a + 5d] + [a + 9d]
⇒ 44 = a + 5d + a + 9d
⇒ 44 = 2a + 14d
⇒ a + 7d = 22
On solving (i) & (ii), we get
⇒ a + 5d = 12
⇒ a + 7d = 22
----------------
-2d = -10
d = 5
Substitute d = 5 in (ii), we get
⇒ a + 7d = 22
⇒ a + 7(5) = 22
⇒ a = 22 - 35
⇒ a = -13.
So,
First term = a₁ = -13
Second term = a₂ = -13 + 5 = -8
Third term = a₃ = -8 + 5 = -3
Therefore, the three terms of the AP are : -13, -8, -3.
Hope it helps!
Answer:
- 13 , - 8 , - 3 .
Step-by-step explanation:
Let the first term a and common difference be d.
We know :
t_n = a + ( n - 1 ) d
t_4 = a + 3 d
t_8 = a + 7 d
We have given :
t_4 + t_8 = 24
2 a + 10 d = 24
a + 5 d = 12
a = 12 - 5 d ....( i )
t_6 = a + 5 d
t_10 = a + 9 d
: t_6 + t_10 = 44
2 a + 14 d = 44
a + 7 d = 22
a = 22 - 7 d ... ( ii )
From ( i ) and ( ii )
12 - 5 d = 22 - 7 d
7 d - 5 d = 22 - 12
2 d = 10
d = 5
We have :
a = 12 - 5 d
a = 12 - 25
a = - 13
Now required answer as :
- 13 , - 8 , - 3 .
Finally we get answer.