Answers

2016-03-25T08:42:05+05:30
Please refer to the image
0
  • qais
  • Content Quality
2016-03-25T08:44:45+05:30
Let first term be a and common difference be d
a1 = a
a5= a+(n-1)d = a +4d
a10 = a +9d
a15= a+14d
a20 = a+19d
a24 = a+23d
now, adding all,
a1+a5+a10+a15+a20+a24 = 225
⇒a + a+4d +a+9d+ a+14d +a+19d +a+23d = 225
⇒6a + 69d = 225
⇒3(2a+23d)= 225
⇒2a +23d = 75______(1)
now,A/q
a1+a2+a3........a24 = (n/2)[2a +(n-1)d]
here, n = 24
(24/2)[2a +(24-1)d]
=12[2a +23d]
from equation (1)
2a +23d = 75
∴12(2a+ 23d) = 12×75 = 900

0