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

 

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

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$\displaystyle\bf\\2)\\A_{61}=a_1+(61-1)\times r=-113+60\times5=-113+300=187\\\\S_{61}=(-113)+(-108)+(-103)+...+(-8)+(-3)+2+7+12+...+187\\\\S_{61}=SA+SB\\\\SA=(-113)+(-108)+(-103)+...+(-8)+(-3)=-(3+8+103+108+113)\\\\n=\frac{113-8}{5}+1=\frac{105}{5}+1=21+1=22~terms\\SA=-\Big( \frac{n(113+3)}{2} \Big)=-\Big( \frac{22\times116}{2} \Big)=-11\times116 =-1276$

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$\displaystyle\bf\\SB=2+7+12+...+187\\\\n=\frac{187-2}{5}+1=\frac{185}{5}+1=37+1=38~terms\\\\SB=\frac{n(187+2)}{2}=\frac{38\times189}{2}=19\times189=3591\\\\S_{61}=SA+SB=-1276+3591=2315\\\\\boxed{\bf S_{61}=2315}$