Question

Which term of the AP 150,147,144.....is it’s first negative term?

Answer 1

Given :

Airthemtic progression series 150,147,144,.....

To find :

First negative term in the series

Theory :

${\purple{\boxed{\large{\bold{a _{n} = a + (n - 1)d}}}}}$

Theory :

Given Ap series:

150,147,144,...

In this series ;

• First term ,a = 150
• common difference ,d = 147-150=-3

we have to find first negative term in the given AP series .

⇒ The value of a+(n-1) d < 0

$\sf\:a+(n-1)d <0$

$\sf\:150+(n-1)(-3)<0$

$\sf150-3n +3<0$

$\sf\:153-3n<0$

$\sf3n>153$

$\sf\:n >\frac{153}{3}$

$\sf\:n>51$

As n is a integer, the first value which satisfies the above condition is n =52

Now,

$\sf\:a _{52} =a + (52- 1)d$

$\sf\:a _{52} =150+ (52- 1)(-3)$

$\sf\:a _{52} =150-156+3$

$\sf\:a _{52} =-3$

Therefore first negative no is -3

$\rule{200}2$

More About Arithmetic Progression:

1) Genral term of an Ap

$\sf\:a_n=a+(n-1)d$

2)Sum of n terms of an AP given by :

$\sf \: S_{n} = \dfrac{1}{2}(2a+ (n - 1)d)$

Answer 2

Answer:

YOUR QUESTION :

Which term of the AP 150,147,144.....is it’s first negative term?

ANSWER :

Given :

Airthemtic progression series :150,147,144.

To find :

First negative term in the series.

Theory :

${\green{\boxed{\large{\bold{a _{n} = a + (n - 1)d}}}}}$

Given Ap series:

150, 147, 144.

First term ,a = 150.

common difference ,d = 147-150=-3.

We have to find first negative term in the given AP series .

=> The value of a+(n-1) d < 0

a+(n-1)d < 0

=> 150 + (n-1)(-3) < 0

=> 150-3n +3 <0

=> 153 - 3n <0

=> 3n>153.

=> n > 153/3

=> n > 51

As n is a integer, the first value which satisfies the above condition is n =52

Now,

$\sf \: a_{52} = a+(6 - 1)d \\ = >\sf \: a_{52} = 150 + (52 - 1)( - 3) \\ = >\sf \: a_{52} = 150 - 156 + 3 \\ = > \sf \: a_{52} = - 3</p><p>$

Therefore first negative no is -3