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1+3+5+7+...+25 is subtracted from 2+4+6+8+...+26. The answer is​

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S_1 = 2+4+6+8+...+26 \\\\~~~~=2(1+2+3+4+...+13)\\\\~~~~=2 \cdot \frac{13(13+1)}2~~~~~~~~;\left[\text{Sum of consecutive positive integers }= \frac{n(n+1)}2 \right] \\\\~~~~=13 \cdot 14\\\\~~~~=182\\\\S_2 = 1+3+5+7+...+25\\\\\text{It is an arithmetic series wihere}\\\\\text{First term, a =1, ~Common difference, d = 2,~ number of terms = n} \\\\~~~~\text{Nth term} = a+(n-1)d\\\\\implies 25 = 1+(n-1)2\\\\\implies 2n-2 = 24\\\\\implies 2n = 24+2\\\\\implies 2n= 26\\\\


\implies n = \frac{26}2 \\\\\implies n= 13\\\\\text{S} _2= \frac{\text{n(First term+ Last term)}}{2}\\\\~~~~~~=\frac{13(1+25)}2\\\\~~~~~~=\frac{13 * 26}2\\\\~~~~~~=13 * 13\\\\~~~~~~=169\\\\\text{Hence,}~~ S_1 - S_2 =182 - 169 =13

User Ibocon
by
8.2k points
9 votes

Answer:

13

Explanation:

2 + 4 + 6 + 8 + ...... + 26

n = 13

Sum of first 'n' even numbers = n( n +1)

= 13 * 14

= 182

1 + 3 + 5 + 7 + .....+25

n = 13

Sum of first 'n' odd numbers = n²

= 13 * 13

= 169

2 + 4 + 6 + 8 + ... + 26 - (1 + 3 + 5 + 7 + ......+25) = 182 - 169

= 13

User Gregor Koukkoullis
by
7.8k points

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