how many terms of the A.P : -15,-11, -7 are needed to make the sum 744
Answers
Answer:
answer below
Step-by-step explanation:
a=-15, d=-11+15=4 , Sn=744
Sn=n/2{2a+(n-1)d} ,
n/2{-30+(n-1)4}=744 ,
n/2(-30+4n-4)=744 ,
n(-34+4n)=1488 ,
4n^2-34n-1488=0 ,
2n^2-17n-744=0 ,
2n^2-48n+31n-744=0 ,
2n(n-24)+31(n-24)=0 ,
(n-24)(2n+31)=0. ,
n=24 or n=-31/2(which is not possible).
Hence,n=24
Answer:
The First 24 terms of the A.P add upto 744.
Step-by-step explanation:
We have,
A.P = (-15), (-11), (-7).........
Now, we need this A.P to add upto 744.
Here,
a = (-15)
d = (-11) - (-15)
= (-11) + 15
= 4
So,
d = 4
Now,
Let till nth term of the A.P, add upto 744.
We know that, Sum of A.P is given by the equation,
Sn = (n/2)[2a + (n - 1)d]
So,
Sn = 744
But we do not know n,
So,
a = (-15)
d = 4
Sn = 744
So,
744 = (n/2)[2(-15) + (n - 1)4]
744 = (n/2)[(-30) + 4n - 4]
744 = (n/2)[4n - 34]
744 = (n/2) × 2[2n - 17]
744 = n(2n - 17)
2n² - 17n = 744
2n² - 17n - 744 = 0
an² + bn + c = 0
where a = 2, b = (-17), c = (-744)
By Splitting the middle term method,
Sum = b = (-17)
Product = (a × c) = (-1488)
So, The Factors are (-48) and 31
Then,
2n² - 48n + 31n - 744 = 0
2n(n - 24) + 31(n - 24) = 0
(n - 24)(2n + 31) = 0
So,
n = 24 or n = (31/2)
Since number of terms will always be a natural number,
n = 24
Hence,
The First 24 terms of the A.P add upto 744.
Hope it helped and believing you understood it........All the best