Question

Find the number of terms in the series 7+11+15+19 ......... of which the term is 250​

Answer 1

Answer:

10

Step-by-step explanation:

7, 11, 15, 19, 23 . . . . . .

In the above AP, first term ‘a’= 7. Common difference ‘d’ = 4, Suppose total ’n' terms make the sum = 250

& Sn = n/2 * [ 2a + ( n-1)*d ] =

=> 250 = n/2 [ 14 + (n-1)*4 ]

=> 500 = n[ 14+4n-4] = n[10+4n]

=> 4n² + 10n - 500 = 0

=> 2n² + 5n - 250 = 0

=> 2n² +25n-20n -250 = 0

=> n(2n+25) - 10 (2n+ 25) = 0

=> ( 2n+ 25) (n-10 ) = 0

=> n = 10 & n= -25/2 ( which is ruled out, as n is no of terms)

=> n = 10

Total 10 terms will make the required sum. please mark brainliest and thank

Answer 2

Answer:

n = 9

Step-by-step explanation:

I guess the number 250 is the sum of all terms in the series.

The series 7+11+15+19.... is an AP

Therefore, a = 7, d = 4, S = 250

S =$\frac{n}{2}$ [2a + (n-1)d]

250 x 2 = n [2x7 + (n-1)4]

500 = n [14 + 4(n-1)]

500 = 14n + 4n(n-1)

500 = 14n + 4$n^{2}$ - 4

500+4 = 14n + 4$n^{2}$ ⇒ 504 = 14n + 4$n^{2}$

4$n^{2}$ + 14n - 504 = 0

Now solving the quadratic equation you get,

n= ≈ (-13) or 9

Since n cannot be in negative, n = 9