Question

how many terms of an AP 36,32,28,24............must be taken so that their sum is 168​

Answer 1

Answer:

12 , 7

Step-by-step explanation:

AP 36,32,28,24.........

d = -4

168 = n{ 72 -4(n-1)}/2

n{ 72 -4(n-1)} = 168x2

n{ 18 -(n-1)} = 84

n(19-n) = 84 = 19n -n²

n² - 19n + 84 = 0

n = 12 , 7

Answer 2

Answer:

the value of n is equal to 4,

there are 4 terms of an AP.

Step-by-step explanation:

The given AP terms in the question are

36, 32,28,24....

a1= 36

Now a2= a1+d=32

a2= 36+d=32

d=32-36,

d= -4

NOW we know that the sum of n terms of an AP is equal to

Sn= n/2{2a+(n-1)d}

Sn=168 given in the question.

Now put the values in the equation we get

168= n/2{2(36) +(n-1) (-4) }

168×2 = n{72+(n-1) (-4) }

336=n{72-4n+4}

336=n(76-4n)

336= 76n-4n^2

-4n^2+76n-336=0

2(2n^2+36n-168) =0

2n^2+36n-168=0

Now multiple first term and last and then Factorise that term. We get

2n^2 ++21n-16n-168=0

n(2n+21) -4(-4n+42) =0

Now (n-4) and (2n+21)

n=4 and n= -21/2

As we know that n cannot never be negative and in fraction So therefore the value of n = 4