No of terms of the sequence -12,-9,-6,-3........ Must be added to make the sum 54 is?????
Answers
Step-by-step explanation:
given ,
general term (a) = -12
common difference (d) = 3
sum of n terms (Sn) of an AP with general term a and common difference d is :
n/2[2a+(n-1)d]
Sn of this AP = n/2[-24 + (n-1)3] n(3n-27)/2
it is given that sum of n terms is 54
so,
n(3n-27)/2 = 54
3n^2 -27n -108=0
n^2 - 9n -36=0
n^2 - 12n +3n -36 =0
n(n-12) +3(n-12)=0
(n+3)(n-12) = 0
n=12. (as n is not a negative number)
so 12 terms of the series should be added to get the sum 54
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Answer:
n = 12
Step-by-step explanation:
First number of the AP = -12
Difference = 3
Sum of 'n' digits = 54
Sn = n/2 [2a+(n-1)d]
Substituting values
54 = n/2 [2(-12)+(n-1)3]
Simplifying the equation
54 = n/2 [-24+3n-3]
54 = n/2 [-27+3n]
54 = -27n+3n^2/2
2*54 = -27n+3n^2
Rearranging the equation
108 = -27n+3n^2
Further simplification (divided by 3) and rearrangement
n^2-9n-36=0
Solving by splitting the middle term method
n^2-12n+3n-36 = 0
n(n-12)+3(n-12) = 0
(n-12)(n+3) = 0
n = 12, 3
(n cannot be a negative value)
Therefore, n = 12