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Given that the age of youngest boy was 8 years and the ages of rest of participants are having a common difference of 4 months. if the sum of ages of all the participants is 168 years.
Let a be the first term and d be the common difference and sn be the sum.
Given first term a = 8 years.
Common difference d = 4 months = 4/12 years
= 1/3 years.
Sn = 168 years.
We know thst sum of n terms of an AP sn = n/2(2a + (n - 1) * d)
168 = n/2(2(8) + (n - 1) * 1/3)
168 * 2 = n(16 + (n - 1)/3)
336 = n(48 + n - 1)/3
336 * 3 = n(47 + n)
1008 = 47n + n^2
n^2 + 47n - 1008 = 0
n^2 + 63n - 16n - 1008 = 0
n(n + 63) - 16(n - 63) = 0
(n + 63)(n - 16) = 0
n = -63 (or) n = 16.
Since n value cannot be negative, so n = 16.
Therefore the age of the eldest participant = a + (16 - 1) * d
= a + 15d
= 8 + 15 * 1/3
= 8 + 5
= 13.
Therefore the age of the eldest participant in the painting competition = 13 years.
Hope this helps!
Let a be the first term and d be the common difference and sn be the sum.
Given first term a = 8 years.
Common difference d = 4 months = 4/12 years
= 1/3 years.
Sn = 168 years.
We know thst sum of n terms of an AP sn = n/2(2a + (n - 1) * d)
168 = n/2(2(8) + (n - 1) * 1/3)
168 * 2 = n(16 + (n - 1)/3)
336 = n(48 + n - 1)/3
336 * 3 = n(47 + n)
1008 = 47n + n^2
n^2 + 47n - 1008 = 0
n^2 + 63n - 16n - 1008 = 0
n(n + 63) - 16(n - 63) = 0
(n + 63)(n - 16) = 0
n = -63 (or) n = 16.
Since n value cannot be negative, so n = 16.
Therefore the age of the eldest participant = a + (16 - 1) * d
= a + 15d
= 8 + 15 * 1/3
= 8 + 5
= 13.
Therefore the age of the eldest participant in the painting competition = 13 years.
Hope this helps!
siddhartharao77:
If possible brainliest it. Thanks.
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