Question

how many terms of the ap 5,7,9 have to be added to get the sum of 320
step by step explanation ​

Answer 1

Answer:

We are given to find x such that

5 + 7 + 9 + . . . . . . + x = 320.

The left hand side of the above equation is an arithmetic series with first term, a = 5 and common difference, d = 7 - 5 = 9 - 7 = . . . = 2.

Also, the sum of first n terms of an arithmetic series with first term a and common difference d is

S_n=\dfrac{n}{2}(2a+(n-1)d).S

n

=

2

n

(2a+(n−1)d).

Let the sum of first n terms be 320. So, we must have

\begin{gathered}\dfrac{n}{2}(2\times5+(n-1)2)=320\\\\\Rightarrow n(5+n-1)=320\\\\\Rightarrow n^2+4n-320=0\\\\\Rightarrow n^2+20n-16n-320=0\\\\\Rightarrow (n-16)(n+20)=0\\\\\Rightarrow n=16,-20.\end{gathered}

2

n

(2×5+(n−1)2)=320

⇒n(5+n−1)=320

⇒n

2

+4n−320=0

⇒n

2

+20n−16n−320=0

⇒(n−16)(n+20)=0

⇒n=16,−20.

Since the number of terms cannot be negative, so n = 16.

Therefore, the sum of first 16 terms is 320 and the 16th term is given by

a_{16}=a+(16-1)d=5+(16-1)\times2=5+30=35.a

16

=a+(16−1)d=5+(16−1)×2=5+30=35.

Thus, the required value of x is 35.