there are five terms in an ap the sum of these terms is 55 and the fourth term is 5 more than the sum of the first two terms find the terms of a p
Answers
Given :
First condition
There are five terms in an ap & the sum of these terms is 55.
Second condition
The fourth term is 5 more than the sum of the first two terms.
To find :
Terms of APs
Solution :
According to the first condition
There are five terms in an ap & the sum of these terms is 55.
As we know that sum of "n" terms of AP is given as :
Sn=2n[2a+(n−1)d]
Where,
a = first term
d = Common difference
⟹S5=25[2a+(5−1)d]=55
⟹25(2a+4d)=55
⟹2a+4d=5511×52
⟹2(a+2d)=22
⟹a+2d=11(equation1)
According to the second condition
The fourth term is 5 more than the sum of the first two terms.
⟹a4=5+S2
⟹a+(4−1)d=5+22[2a+(2−1)d]
⟹a+3d=5+2a+d
⟹a−2a+3d−d=5
⟹−a+2d=5(equation2)
Add both the equations
→ a + 2d + (-a + 2d) = 11 + 5
→ a + 2d - a + 2d = 16
→ 4d = 16
→ d = 16/4 = 4
Put the value of d in eqⁿ 1
→ -a + 2d = 5
→ -a + 2 × 4 = 5
→ -a + 8 = 5
→ a = 8 - 5 = 3
Terms of AP
→ a₁ = 3
→ a₂ = a + d = 3 + 4 = 7
→ a₃ = a + 2d = 3 + 8 = 11
→ a₄ = a + 3d = 3 + 12 = 15
→ a₅ = a + 4d = 3 + 16 = 19
•°• Required A.P = 3, 7, 11, 15, 19..
Given :
First condition
There are five terms in an ap & the sum of these terms is 55.
Second condition
The fourth term is 5 more than the sum of the first two terms.
To find :
Terms of APs
Solution :
According to the first condition
There are five terms in an ap & the sum of these terms is 55.
As we know that sum of "n" terms of AP is given as :
Sn=2n[2a+(n−1)d]
Where,
a = first term
d = Common difference
⟹S5=25[2a+(5−1)d]=55
⟹25(2a+4d)=55
⟹2a+4d=5511×52
⟹2(a+2d)=22
⟹a+2d=11(equation1)
According to the second condition
The fourth term is 5 more than the sum of the first two terms.
⟹a4=5+S2
⟹a+(4−1)d=5+22[2a+(2−1)d]
⟹a+3d=5+2a+d
⟹a−2a+3d−d=5
⟹−a+2d=5(equation2)
Add both the equations
→ a + 2d + (-a + 2d) = 11 + 5
→ a + 2d - a + 2d = 16
→ 4d = 16
→ d = 16/4 = 4
Put the value of d in eqⁿ 1
→ -a + 2d = 5
→ -a + 2 × 4 = 5
→ -a + 8 = 5
→ a = 8 - 5 = 3
Terms of AP
→ a₁ = 3
→ a₂ = a + d = 3 + 4 = 7
→ a₃ = a + 2d = 3 + 8 = 11
→ a₄ = a + 3d = 3 + 12 = 15
→ a₅ = a + 4d = 3 + 16 = 19
•°• Required A.P = 3, 7, 11, 15, 19..