A question on arithematic Proggression. Please help !?

Question

I'am preparing for exams so please help me people
the question is :
If a1, a2, a3, ........ are in A.P. such that a1+a5+ a10+ a15+ a20+ a24+ =225 then find the value of a1+ a2+ a3+ a4+ ........+a24.
Please also show me how you got the answer.

Answer 1

The sum of an a.p. is n * average of first and nth term,
i.e. n * [ ( a + a + (n-1) * d ) / 2 ], where a+(n-1)d is the last term
Observe that a1 + a24 = a5 + a20 = a10 + a15 ( = 2a1 + 23d )
Thus we have 3 * (a1 + a24) = 225 => a1 + a24 = 225/3 = 75.
Thus the sum of the a.p upto 24 terms is
24 * (75/2) = 900

Answer 2

First off, a[1] is equal to some value a.

a[1] = a

a[n] = a + (n - 1)d

Therefore,

a[1] + a[5] + a[10] + a[15] + a[20] + a[24] = 25
a + (a + 4d) + (a + 9d) + (a + 14d) + (a + 23d) = 225

5a + 50d = 225

We can reduce this by dividing both sides by 5. This gives us

a + 10d = 45

All this tells us is that the 9th term, a[9], is 45.
I think there's not enough information.