How many terms of the A.P:24,21,18,........multiple be taken so that their sum is 78?

How many terms of the A.P:24,21,18,........multiple be taken so that t

1.	How many terms of the A.P:24,21,18,........multiple be taken so that their sum is 78?
Answer» Here a = 24, d = 21\xa0- 18\xa0= - 3, Sn\xa0= 78Sn= {tex}\\frac n2{/tex}[2a + (n - 1)d]{tex}\\therefore \\frac n2{/tex}[2(24) + (n - 1)(-3 )] = 78{tex}\\therefore \\frac n2{/tex}[48\xa0+ (n - 1)(-3 )] = 78or, n(48\xa0-3 n+ 3) = 0or, n(- 3n + 48) = 0or, n = 0 and -3n + 48\xa0= 0-3n = -48n = 48/3 = 16n = 16