Q. 22

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Answer :

Given:


Age of youngest boy = 8 years


Common difference between ages = 4 months years


We can consider ages of the students in AP with


First term, a = 8 years


Common difference,


Let the number of children be ‘n’.


As, sum of ages = 168 year


Sn = 168


Also, we know sum of n terms of an AP



Where,


a = first term


d = common difference


n = number of terms


Putting values, we get


Sn = 168





n(47 + n) = 1008


n2 + 47n – 1008 = 0


n2 + 63n – 16n – 1008 = 0


n(n + 63) – 16(n + 63) = 0


(n + 63)(n – 16) = 0


n = 16 or n = –63 [Not possible, as number of students can’t be negative]


Also, we know nth term of an AP


an = a + (n – 1)d


a16


a16 = 13 years


Hence, age of eldest student is 13 years.


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