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if T5=>11, T9=> 7, Then T16=> ?
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Hello there!
Topic: Arithmetic Progression.
Thank you!
===================================
Given Data:
T₅ = 11
T₉ = 7
T₁₆ = ? (to be find)
Formula going to be used: (Arithmetic Progressions)
To find nth number of the series, the formula is Tₓ = a + (n-1) d
where,
a is the first term of the series
d is the common difference between two consecutive terms.
n is the position of the term.
Solution:
Using Tₓ = a + (n-1) d, we will obtain two equation by following the data given.
Case 1: T₅ = a + (5-1)d [as T₅=11]
11 = a + 4d. or a + 4d = 11 --------------- 1
Case 2: T₉ = a + (9-1)d [as T₉=7]
7 = a+8d or a+8d = 7 ---------------- 2
Solving equations 1 and 2, we get
[ multiply eqn 2 with minus for easy solving]
a + 4d = 11
-a - 8d = -7
_________
- 4d = 4
_________
d = -1 [common difference]
Substitute the d value in eqn 1 or 2:
a + 4(-1) = 11
a = 11+4 = 15
a = 15 [starting value]
To find T₁₆ :
T₁₆ = a + (16-1)d
= 15 + 15(-1)
= 0
Therefore, 16th term of the arithmetic series (T₁₆) is 0.
Topic: Arithmetic Progression.
Thank you!
===================================
Given Data:
T₅ = 11
T₉ = 7
T₁₆ = ? (to be find)
Formula going to be used: (Arithmetic Progressions)
To find nth number of the series, the formula is Tₓ = a + (n-1) d
where,
a is the first term of the series
d is the common difference between two consecutive terms.
n is the position of the term.
Solution:
Using Tₓ = a + (n-1) d, we will obtain two equation by following the data given.
Case 1: T₅ = a + (5-1)d [as T₅=11]
11 = a + 4d. or a + 4d = 11 --------------- 1
Case 2: T₉ = a + (9-1)d [as T₉=7]
7 = a+8d or a+8d = 7 ---------------- 2
Solving equations 1 and 2, we get
[ multiply eqn 2 with minus for easy solving]
a + 4d = 11
-a - 8d = -7
_________
- 4d = 4
_________
d = -1 [common difference]
Substitute the d value in eqn 1 or 2:
a + 4(-1) = 11
a = 11+4 = 15
a = 15 [starting value]
To find T₁₆ :
T₁₆ = a + (16-1)d
= 15 + 15(-1)
= 0
Therefore, 16th term of the arithmetic series (T₁₆) is 0.
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