Question

7. In the arithmetic sequence 60, 56, 52, 48,g , starting from the first term, how many
terms are needed so that their sum is 368?

Answer 1

In the attachments I have answered this problem.

The sum of n terms of an AP is

S(n) = (n/2) [2a + (n-1) d]

This formula has been applied to find n.

See the attachments for detailed solution

I hope this answer helps you

Answer 2

SOLUTION :

Given :
S n = 368
60,56,52,48,.......
a = 60 , d = 56 - 60 = -4

Sum of n terms, Sn = (n/2) [ 2a + (n-1) d ]
Sn= (n/2) [ 2(60) + (n-1)(-4) ]
368 = (n/2) [ 120 - 4 n + 4 ]
368 = (n/2) [ 124 - 4 n ]
368 x 2 = n [ 124 - 4 n ]
368 x 2 = n [ 124 - 4 n ]
124 n - 4 n² = 736
4 n² - 124 n + 736 = 0
4( n² - 31 n + 184)= 0
n² - 31 n + 184= 0
n² -23n - 8n + 184 = 0
[ By middle term splitting]
n(n -23) -8(n -23) = 0
(n-23) (n-8) = 0
n = 23,8

Hence, 8 or 23 terms are needed

HOPE THIS WILL HELP YOU...