Question

first term of arithmetic sequence is 1 and the sum of the first 4 terms is 100 find the first 4 terms

Answer 1

Note : This isn't a Biology Question, This question comes under Mathematics,

Here is the solution :

Given that,
First term of A.S = 1,

In general we consider a as the first term,

So, Let a = 1,

Let the next other 3 terms be , b,c,d

According to the Question,

a + b + c + d = 100,

We know the Formula for Sum of n terms of Arthimetic Sequence,

Sn(Sum of n terms) = (n/2)(a+l)

Where a = first term, l = last term, n = no.of terms,

=> 100 = (4/2)(1 + l)

=> 50 = 1 + l,

=> l = 49,

Now, We have to find Common difference(d),

Nth term of A.P = a + (n-1)*d,

=> 49 = 1 + 3*d,

=> 48 = 3d,

=> 16 = d,

=> b = 1 + 16 = 17,

=> c = 17 + 16 = 33,

=> d = 33 + 16 = 49 ✔(This is what we got before)

Therefore : First 4 terms of the A.P are 1 , 17 , 33 , 49 ,

Hope you understand, Have a Great day :D,
Thanking you, Bunti 360 !..

Answer 2

Let a, a+d, a+2d, a+3d be the 4 terms of Ap a=1, a+(a+d) +(a+2d) +(a+3d) =100
1+(1+d)+(1+2d)+(1+3d)=100
1+1+d+1+2d+1+3d=100
4+6d=100
6d=100-4
6d=96
d=96÷6
d=16
a=1,d=16
Ap=a,a+d,a+2d,a+3d
Ap=1,1+16,1+2(16),1+3(16)
Ap=1,17,1+32,1+48
Ap=1,17,33,49