How many terms of the A.P. 16, 14, 12,.. are needed to give the sum 60? Explain
why do we get two answers.
Answers
First term(a) = 16
Common difference(d) = 14 - 16 = - 2
Sum of term(S) = 60
Let n terms are required.
Using, S = (n/2) [2a + (n - 1)d]
=> 60 = (n/2) [2(16) + (n - 1)(-2)]
=> 2 × 60 = n [32 - 2n + 2]
=> 2 × 60 = n(34 - 2n)
=> 60 = n(17 - n)
=> 60 = 17n - n^2
=> n^2 - 17n + 60 = 0
=> n^2 - (12 + 5)n + 60 = 0
=> n^2 - 12n - 5n + 60 = 0
=> n(n - 12) - 5(n - 12) = 0
=> (n - 12)(n - 5) = 0
=> n = 12 or n = 5
Notice that the common difference is negative because of which it can easily be said that upcoming terms would be negative and would cancel out many or all positive terms. The same happened in this case that for n = 5, +ve terms add upto 60. But for n = 12, there comes few more +ve values but the equal -ve values which cancel out each other. It is like '1 + 2 = 3' also '1 + 2 + 3 + (-3) = 3' (just an example, not AP).
n = 5 or n = 12.
Answer:
n = 5 or 12
5 or 12 terms of the A.P : 16, 14, 12,.. are needed to give the sum 60.
Step-by-step explanation:
Given :-
A.P = 16, 14, 12,..
To find :-
How many terms are needed to give the sum 60.
Solution :-
We have,
Fist term a = 16
Common difference d = 14 - 16 = 12 - 14 = -2
Sum of terms s = 60
Using the formula,
Sum of first n terms in an A.P.
- Sₙ = n/2 (2a + (n-1)d
Substituting we get,
60 = n/2 (2 × 16 + (n-1)-2
120 = n(34-2n)
120 = 34 - 2n²
n² - 17n + 60 = 0
12 × 5 = 60 -12 - 5 = -17
n² - 12n - 5n + 60 = 0
n(n - 12) - 5(n - 12) = 0
n - 5 = 0 & n - 12 = 0
∴ n = 5 or 12
Hence we have, S₅ = 78 and S₁₂ = 78 and all the terms from S₅ to S₁₂ are 0.