Question

how many terms of the AP 45 39,33 must be taken so that there is sum is 180 ​

Answer 1

SOLUTION:-

Given:

AP is 45,39,33.......

Using Formula of the sum:

$Sum = \frac{n}{2} [2a + (n - 1)d]$

Here,

First term(a)= 45

Common difference= 39-45= -6

Sum= Sn= 180

We have to find the value of n:

$Sum = \frac{n}{2} [2a + (n - 1)d] \\ \\ = > 180 = \frac{n}{2} [2 \times 45 + (n - 1)( - 6)] \\ \\ = > 180 \times 2 = n[90 + (n - 1)( - 6)]\\ \\ = > 360 = n[90 - 6n + 6] \\ \\ = > 360 = n[96 - 6n]\\ \\ = > 360 = 96n - 6 {n}^{2} \\ \\ = > 6 {n}^{2} - 96n + 360 = 0$

Dividing the equation by 6:

$= > \frac{6 {n}^{2} }{6} - \frac{96n}{6} + \frac{360}{6} = 0 \\ \\ = > {n}^{2} - 16n+ 60 = 0$

Now,

Factorize by splitting the method term:

n² -16n + 60 =0

=) n² -10n-6n +60=0

=) n(n-10)-6(n-10)=0

=) (n-10)(n-6)=0

=) n-10= 0 or n-6= 0

=) n= 10 or n= 6

Since,

Both values are positive natural number.

Hope it helps ☺️

Answer 2

Answer:

Step-by-step explanation:

SOLUTION:-

Given:

AP is 45,39,33.......

Using Formula of the sum:

Here,

First term(a)= 45

Common difference= 39-45= -6

Sum= Sn= 180

We have to find the value of n:

Dividing the equation by 6:

Now,

Factorize by splitting the method term:

n² -16n + 60 =0

=) n² -10n-6n +60=0

=) n(n-10)-6(n-10)=0

=) (n-10)(n-6)=0

=) n-10= 0 or n-6= 0

=) n= 10 or n= 6

Since,

Both values are positive natural number.