Question

помогите с заданнием: 105, 107, 109​

Answer 1

$\displaystyle\bf\\105)\\\\a_{8} ^{2} -a_{2} ^{2} =912\\\\(a_{1} +7d)^{2} -(a_{1} +d)^{2} =912\\\\a_{1} ^{2} +14a_{1} d+49d^{2} -a_{1} ^{2} -2a_{1} d-d^{2}=912\\\\12a_{1} d+48d^{2} =912\\\\a_{1} d+4d^{2} =76\\\\d\cdot\underbrace{(a_{1} +4d)}_{a_{5}=19 }=76\\\\d=76:19=4\\\\a_{1} =19-4d=19-4\cdot 4=3\\\\a_{2} =a_{1}+d=3+4=7\\\\Otvet \ : \ a_{2} =7$

$\displaystyle\bf\\107)\\\\S_{15} -S_{13} =16\\\\\\\frac{2a_{1} +14d}{2} \cdot 15-\frac{2a_{1} +12d}{2} \cdot 13=16\\\\\\(a_{1} +7d)\cdot 15-(a_{1} +6d)\cdot 13=16\\\\15a_{1} +105d-13a_{1} -78d=16\\\\\boxed{2a_{1} +27d=16}\\\\\\S_{28} =\frac{2a_{1}+27d }{2}\cdot28=\underbrace{(2a_{1} +27d)}_{16}\cdot 14=16\cdot 14=224\\\\Otvet \ : \ S_{28} =224$

$\displaystyle\bf\\109)\\\\d=\frac{a_{20} -a_{2} }{a_{13} -a_{4} } =\frac{a_{1}+19d-a_{1}-d }{a_{1} +12d-a_{1} -3d} =\frac{18d}{9d} =2\\\\\\\boxed{d=2}\\\\a_{3} =5\\\\a_{1} +2d=5\\\\a_{1} =5-2d=5-2\cdot 2=5-4=1\\\\\boxed{a_{1} =1}\\\\a_{7} =a_{1} +6d=1+6\cdot 2=1+12=13\\\\Otvet \ : \ a_{7} =13$