Question

How many terms of the arithmetic series 24+21+18+15............be taken continuously,so that their sum is -351?

Answer 1

Hey sup!

As per the question,

a=24.
d=21-24= -3.

=>S(n)=-351.
=>n/2{2a+(n-1)d}=-351.
=>n/2{2*24+(n-1)-3=-351.
=>n/2(48-3n+3)=-351.
=>n/2(51-3n)=-351.
=>n(51-3n)=-351*2.
=>51n-3n^2=-702.
=>51n-3n^2+702=0.
=>-3(n^2-17n-234)=0.
=>n^2-17n-234=0/-3=0.
=>n^2-26n+9n-234=0.
=>n(n-26)+9(n-26)=0.
=>(n-26)((n+9)=0.
We have got two roots.
We'll discard n+9 as it gives-ve value.

So, n-26=0.
=>n=26.

So,26 terms of AP series 24+18+15+... Will have to be taken continuously so that their sum is -351.

Hope it helps.

Answer 2

Answer:

26 terms

Step-by-step explanation:

AP is 24, 21, 18,...

Sn = -351, n = ?

Given a = 24,

d= 21 - 24 = -3

Now, Sn = n[2a + (n-1)d]/2

∴ -351 = n[2x24 +(n-1)(-3)]

-351 x 2 = n(48 -3n + 3) = n(51 - 3n)

-702 = 51n - 3n²

3n² - 51n - 702 = 0

Dividing by 3

n² - 17n - 234 = 0

n² - 26n + 9n - 234 = 0

n(n-26) + 9(n-26) = 0

(n + 9)(n-26) = 0

n = -9 OR n = 26

Discarding negative value of n, we get

n = 26