Q. 22

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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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