c) How many terms of the AP 7, 11, 15, 19, 23,
the sum 250 ?
Answers
QUESTION:
How many terms of an arithmetic sequence 7,11,15,… must be added to get sum 250?
Solution:
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.
Step-by-step explanation:
a = 7
d = a2 - a1
= 11 - 7
= 4
Sn = 250
Sn = n/2 x (2a + (n - 1)d)
250 = n/2 (2(7) + (n - 1)(4))
500 = n(14 + (n - 1)4)
4n² + 10n - 500 = 0
2n² + 5n - 250 = 0
2n² + 25n - 20n - 250 = 0
n(2n + 25) - 10(2n + 25) = 0
(n - 10)(2n + 25) = 0
n - 10 = 0 2n + 25 = 0
n = 10 2n = -25
n = 10 n = -25/2
Therefore n = 10