How many term of arithmetic progression 45, 39, 33 must be taken so that their sum is 180 explain the double answer
Answers
Answered by
105
Hello there ...
here a = 45 , d = -6 , Sn = 180
To find : n
We know ,
Therefore,
180 = n / 2 × 2 ( 45) + (n - 1 ) -6
180 = n/2 × 90 - 6n + 6
180 × 2 = n ( 96 - 6n )
360 = 96n - 6n^2
6n^2 - 96n + 360 = 0
Dividing by 6
n^2 - 16n + 60 = 0
n^2 - 10n - 6n + 60 = 0 .......( factorizing )
n ( n - 10 ) - 6 ( n - 10 ) = 0 .......{ taking common out }
( n - 6 ) (n - 10 ) = 0
n - 6 = 0 or. n - 10 = 0
n = 6. or. n = 10
Therefore 6 or 10 terms must be taken so that their sum is 180
Answered by
42
The number of arithmetic progression are "6 and 10".
Explanation:
Here, first term (a) = 45 , common difference (d) = 39 - 45 = - 6 and
Let the number of term = n
To find, the umber of term = ?
We Know that,
⇒
⇒
⇒
⇒ [divided by 6]
⇒
⇒
⇒
⇒ n = 6 or 10
∴ n = 6 or 10
Hence, the number of arithmetic progression are 6 and 10.
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