The sum

The sum of the first 10 members of the arithmetic sequence is 120. What will be the sum if the difference is reduced by 2?

Result

s2 =  30

Solution:

s1=120 n=10  s1=a1+a2+a3+...+an  d2=d12  s1=10/2 (a1+a1+9 d1)=10 a1+45 d1=120 s2=10/2 [a1+a1+9 (d12)] s2=10 a1+45 d15 9 2  s2=s15 9 2=1205 9 2=30s_{1}=120 \ \\ n=10 \ \\ \ \\ s_{1}=a_{1}+a_{2}+a_{3}+... + a_n \ \\ \ \\ d_{2}=d_{1} - 2 \ \\ \ \\ s_{1}=10/2 \cdot \ (a_{1} + a_{1} + 9 \cdot \ d_{1})=10 \cdot \ a_{1} + 45 \cdot \ d_{1}=120 \ \\ s_{2}=10/2 \cdot \ [a_{1} + a_{1} + 9 \cdot \ (d_{1}-2)] \ \\ s_{2}=10 \cdot \ a_{1} + 45 \cdot \ d_{1} - 5 \cdot \ 9 \cdot \ 2 \ \\ \ \\ s_{2}=s_{1} - 5 \cdot \ 9 \cdot \ 2=120 - 5 \cdot \ 9 \cdot \ 2=30



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