Question

For an arithmetic sequence, a35=57. If the common difference is 5. Find a1 and the sum of the first 61 terms.

Answer 1

`\displaystyle\bf\\a_(35)=57\\r=5\\----\\a_1=?\\S_(61)=?\\----\\Solve:\\1)\\a_1=a_(35)-(35-1)* r=57-34* 5=57-170=-113\\\boxed{\bf a_1=-113}\\\\`

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`\displaystyle\bf\\2)\\A_(61)=a_1+(61-1)* r=-113+60*5=-113+300=187\\\\S_(61)=(-113)+(-108)+(-103)+...+(-8)+(-3)+2+7+12+...+187\\\\S_(61)=SA+SB\\\\SA=(-113)+(-108)+(-103)+...+(-8)+(-3)=-(3+8+103+108+113)\\\\n=(113-8)/(5)+1=(105)/(5)+1=21+1=22~terms\\SA=-\Big( (n(113+3))/(2) \Big)=-\Big( (22*116)/(2) \Big)=-11*116 =-1276`

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`\displaystyle\bf\\SB=2+7+12+...+187\\\\n=(187-2)/(5)+1=(185)/(5)+1=37+1=38~terms\\\\SB=(n(187+2))/(2)=(38*189)/(2)=19*189=3591\\\\S_(61)=SA+SB=-1276+3591=2315\\\\\boxed{\bf S_(61)=2315}`