how many terms of the AP 17,15,13,11,......... must be added to get the sum 72?explain the double answer.
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16
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ѕolυтιon
a = 17
d = 15- 17 =-2
n=?
ѕn=72
==>72= n/2(2*17 + (n-1)*-2)
==>144 = n(34-2n+2)
==>144= 36n - 2n^2
==>n^2-18n+72=0
==>n^2-12n-6n+72
==>n(n-12)-6(n-12)
==>n=12 oғ n=6 anѕ......✳️✳️
ѕolυтιon
a = 17
d = 15- 17 =-2
n=?
ѕn=72
==>72= n/2(2*17 + (n-1)*-2)
==>144 = n(34-2n+2)
==>144= 36n - 2n^2
==>n^2-18n+72=0
==>n^2-12n-6n+72
==>n(n-12)-6(n-12)
==>n=12 oғ n=6 anѕ......✳️✳️
anugyasoni:
what is the explain the double anser
Answered by
20
♧♧HERE IS YOUR ANSWER♧♧
The AP is
17, 15, 13, 11, ...
So, first term, a = 17,
second term = 15.
Then, common difference,
d = 15 - 17 = - 2.
Let us consider the AP contains n number of terms.
Sum of an AP series = 72
=> (n/2)[2a + (n-1)d] = 72
=> (n/2)[2×17 + (n-1)×(-2)] = 72
=> (n/2)[34 - 2n + 2] = 72
=> (n/2)(36 - 2n) = 72
=> n(18 - n) = 72
=> n² - 18n + 72 = 0
=> n² - 12n - 6n + 72 = 0
=> n(n - 12) - 6(n - 12) = 0
=> (n - 12)(n - 6) = 0
So, n = 12, 6.
Here, n = 6 is the required number of terms.
Since the AP is having first term 17 (positive) and common difference (-2) (negative), there are two values of n.
The sum of the last six terms (when n = 12) is 0 indeed.
So, only (6 - 4) = 2 or (12 - 4) = 8 terms need to be added.
♧♧HOPE THIS HELPS YOU♧♧
The AP is
17, 15, 13, 11, ...
So, first term, a = 17,
second term = 15.
Then, common difference,
d = 15 - 17 = - 2.
Let us consider the AP contains n number of terms.
Sum of an AP series = 72
=> (n/2)[2a + (n-1)d] = 72
=> (n/2)[2×17 + (n-1)×(-2)] = 72
=> (n/2)[34 - 2n + 2] = 72
=> (n/2)(36 - 2n) = 72
=> n(18 - n) = 72
=> n² - 18n + 72 = 0
=> n² - 12n - 6n + 72 = 0
=> n(n - 12) - 6(n - 12) = 0
=> (n - 12)(n - 6) = 0
So, n = 12, 6.
Here, n = 6 is the required number of terms.
Since the AP is having first term 17 (positive) and common difference (-2) (negative), there are two values of n.
The sum of the last six terms (when n = 12) is 0 indeed.
So, only (6 - 4) = 2 or (12 - 4) = 8 terms need to be added.
♧♧HOPE THIS HELPS YOU♧♧
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