Question

A club consists of members whose ages are in A.P. , the common differencebeing 3 months. If the youngst member of the club is just 7 years old and sum of the ages of members is 250 years find the number of member in club ?????????????????????

Answer 1

Hey sup!

As per the question,
d=3(months)
a=7(years)=7×12=84(months).
S(n)=250(years)=250*12=3000(months)

S(n)=n/2(2a+(n-1)d).
=>3000=n/2(2*84+(n-1)3.
=>3000*2=n(168+3n-3).
=>6000=n(165+3n).
=>6000=165n+3n^2.
=>3n^2+165n-6000=0.
=>3(n^2+55n-2000)=0.
=>n^2+55n-2000=0/3=0.
=>n^2+80n-25n-2000=0.
=>n(n+80)-25(n+80)=0.
=>(n-25)(n+80)=0.
We'll discard (n+80) as it gives-ve value.
n-25=0.
=>n=25.

There are 25 members in the club.

Hope it helps.

Answer 2

Answer:

25

Step-by-step explanation:

Given :

The common difference =3 months.

The youngest member of the club is just 7 years old i.e. $a_1=7$

Sum of the ages of members is 250

To Find: find the number of member in club .

Solution :

First term of A.P. = 7 years = 7*12=  84 months

Sum of the ages of members is 250  years = 250*12 = 3000 months.

Sum of nth term : $\frac{n}{2}(2a+(n-1)d)$

Where a is the first term

d is the common difference

Given a = 84

d = 3

Sn = 3000

So,  $3000=\frac{n}{2}(2\times 84+(n-1)3)$

$6000=n(168+(n-1)3)$

$6000=n(168+3n-3)$

$6000=n(165+3n)$

$6000=165n+3n^2$

$165n+3n^2-6000=0$

$n^2+55n-2000=0$

$n^2+80n-25n-2000=0.$

$n(n+80)-25(n+80)=0$

$(n-25)(n+80)=0$

So, n = 25 , -80

Since the number of members cannot be negative .

So, total of members in the club is 25