The 14th term of an AP is twice its 18th term. if the 6th term is -8, then find the sum of its first 20 terms
Answers
Answer :
S(20) = -115
Note :
★ A.P. (Arithmetic Progression) : A sequence in which the difference between the consecutive terms are equal is said to be in A.P.
★ If a1 , a2 , a3 , . . . , an are in AP , then
a2 - a1 = a3 - a2 = a4 - a3 = . . .
★ The common difference of an AP is given by ; d = a(n) - a(n-1) .
★ The nth term of an AP is given by ;
a(n) = a + (n - 1)d .
★ If a , b , c are in AP , then 2b = a + c .
★ The sum of nth terms of an AP is given by ; S(n) = (n/2)×[ 2a + (n - 1)d ] .
or S(n) = (n/2)×(a + l) , l is the last term .
★ The nth term of an AP can be also given by ; a(n) = S(n) - S(n-1) .
★ A linear polynomial in variable n always represents the nth term of an AP .
★ A quadratic polynomial in variable n always represents the sum of n terms of an AP .
Solution :
- Given : a(14) = 2•a(18) , a(6) = -8
- To find : S(20) = ?
We have ;
=> a(6) = -8
=> a + (6 - 1)d = -8
=> a + 5d = -8 --------(1)
Also ,
We have ;
=> a(14) = 2•a(18)
=> a + (14 - 1)d = 2•[ a + (18 - 1)d ]
=> a + 13d = 2•(a + 17d)
=> a + 13d = 2a + 34d
=> 2a - a + 34d - 13d = 0
=> a + 21d = 0 --------(2)
Now ,
Subtracting eq-(1) from eq-(2) , we have ;
=> (a + 21d) - (a + 5d) = 0 - (-8)
=> a + 21d - a - 5d = 8
=> 16d = 8
=> d = 8/16
=> d = ½
Now ,
Putting d = ½ in eq-(2) , we have ;
=> a + 21d = 0
=> a = -21d
=> a = -21•½
=> a = -21/2
=> a = -10½
Also ,
We know that , S(n) = (n/2)×[ 2a + (n - 1)d ]
Thus ,
=> S(20) = (20/2)×[ 2a + (20 - 1)d ]
=> S(20) = 10×[ 2a + 19d ]
=> S(20) = 10×[ 2•(-21/2) + 19•(1/2) ]
=> S(20) = 10×[ -21 + 19/2 ]
=> S(20) = -210 + 95
=> S(20) = - 115
Hence , S(20) = -115 .
Given:
- 14th term = 2 × 18th term
- 6th term = a₆ = -8
To find:
- Sum of first 20 terms = S₂₀
Method:
★ We know than aₙ = a + (n - 1)d ★
a₁₄ = a + (14 - 1) d
⇒ a₁₄ = a + 13d
Similarly
a₁₈ = a + 17d
Also,
a₆ = a + 5d = -8 ------ (Equation 1)
According to the Question,
a₁₄ = 2 (a₁₈)
⇒ a + 13d = 2(a + 17d)
⇒ a + 13d = 2a + 34d
⇒ 2a - a + 34d - 13d = 0
⇒ a + 21d = 0 ------ (Equation 2)
Subtracting equations 1 from 2,
a + 21d = 0
(-) a + 5d = -8
16d = 8
⇒ d = 8 ÷ 16
⇒ d = 0.5
Now let us find a
a + 5d = -8
⇒ a + 2.5 = -8
⇒ a = -10.5
For finding the sum of 20 terms,
S₂₀ = ⁿ/₂ [2a + (n - 1)d]
⇒ S₂₀ = 20 ÷ 2 [2 × -10.5 + (20 - 1) 0.5]
⇒ S₂₀ = 10 [-21 + (19) × 0.5]
⇒ S₂₀ = 10 [-21 + 9.5]
⇒ S₂₀ = 10[-11.5]
⇒ S₂₀ = -115
Therefore the sum of the first 20th term is -115