Question

a1, a2, a3,.......a24 are in A.P. and a1+a5+a10+a15+a20+a24=300. find the sum of first 24 terms of A.P.

Answer 1

a1 + a5 + a10 + a15 + a20 + a24 = 300.............eq.1
an = a + (n-1)d
a5 = a +4d
a10 = a+9d
a15 = a+14d 
a20 = a+19d
a24 = a+23d
put the values of a5 a10 a15 a20 a24 in eq.1
a + a + 4d+a+9d+a+14d+a+19d+a+23d= 300
6a + 69d = 300
divide the whole equation from 3 
2a + 23d = 100..........eq.2
Sn = n/2(2a + (n-1)d)
S24 = 24/2(2a + 23d)
from eq.2 
S24 = 12(100)
S24 = 1200

Answer 2

Answer:

900

Step-by-step explanation:

a1+a5+a10+a15+a20+a24=225

here according to arithmetic progression property.

a1+a24=a5+a20=a10+a+a15=225

3(a1+a24)=225

a1+a24=75

sn=n/2[a+l]

a24=12*(a1+a24).

=12*75

=900

hope this is useful.