# For the series 3, 8 15, 24 , 35… what is the difference of the sum of the first 25 terms and the first 75 terms.

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We see that the series 3, 8, 15, 24, 35… is not an AP as there is no common difference; neither is the common ratio a constant.

So looking at other alternatives, we see that each of the terms can be written as a square minus 1

0= 1-1

3=4-1

8= 9-1…

So the sum of the first 25 terms is 1 + 4 + 9 + 16 +… -25*1

The sum of the first 75 terms is 1+ 4+ 9+ 16 +… - 75*1

Now we also know that the sum of the first n squares is given by n*(n+1)*(2n+1)/6

Therefore we require 75*(75+1)*(150+1)/6 – 75 – [25*(26)*(50 + 1)/6 – 25]

=> 75*76*151/6 – 75 – 25*26*51/6 +25

=> 137875

**The required difference of the sum of the first 75 terms and the sum of the first 25 terms is 137875**

The terms of the given series is 3,8,15,24, 35....

a1 = 3, a2 = 8, a3 = 15, a4 = 24, a5 =35

Therefore we notice that an = (n+1)^2-1 = n^2+2n = n(n+1)+n.

Therefore the sum Sn of the n terms of the series could be written as:

Sn = (1*2+1)+(2*2+2) +(3*4+3)+(4*5+4)+(5*6+4)+.....

Sn = sum (n)(n+1) + sum of (1+2+3+..)

Sn = sum n(n+1) + n(n+1)/2

Sn = (n+2)(n+1)(n)/3 +n(n+1)/2

Sn = n(n+1){2(n+2)+3}/6 .

Sn = n(n+1)(2n+7)/6.

Tally: S1 = 1*(1+1)(2*1+7) / 6 = 18/6 = 3. Also S1 =a1 = 3.

S2 = 2(2+1)(2*2+7)/6 = 11. Also S2 = a1+ a2 = 3+ 8 = 11.

S3 = 3(3+1)(2*3+7)/6 = 26, Also S3 = a1+a2+a3 = 3+8+15 = 26.

Therefore the sum of 25 terms S25 = 25(26)(2*25+7)/6 = 6175.

Similarly S75 = (75)(76)(2*75)/6 = 149150.

Therefore S75 - S25 = 149150-6175 = 142975.

Therefore the difference of the sum of the first 25 terms and the sum of the first 75 tems of the series is S75 - S25 = 142975.