✏ FIND THE SUM
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As we know general formula for finding out sum of first n terms of arithmetic progression series is :
[tex]\\[/tex]
[tex]\tt{S_n= \frac{n}{2} *(2a+(n-1) d)}[/tex]
[tex]\\[/tex]
Here, S_25 = - 450, S_56 = 1596 and we have to find S_100
[tex]\\[/tex]
Now from above formula;
[tex]\\[/tex]
[tex]\tt{S_{25}= 25/2 * (2a+(n-1)d)}[/tex]
[tex] \tt∴ -450 = 25/2 * (2a+(25-1)d) [/tex]
[tex] \tt∴ -450 = 25/2 * (2a+24d)[/tex]
[tex]\tt∴ -450 = 25 (a+12d)[/tex]
[tex]\tt∴ a+12d = -450/25 [/tex]
[tex]\tt∴ a+12d+18 = 0 --- EQ (1) [/tex]
[tex]\\[/tex]
Now,
[tex]\\[/tex]
[tex]\tt{S_{56}=56/2 * (2a+(56-1)d)}[/tex]
[tex]\tt∴ 1596 = 28 * (2a+55d) [/tex]
[tex]\tt∴ 2a+55d = 1596/28 [/tex]
[tex]\tt∴ 2a+55d = 57 [/tex]
[tex]\tt∴ 2a+55d-57 = 0 --- EQ (2) [/tex]
[tex]\\[/tex]
Here, we have got two linear equations by solving those equations we can get value of a and d.
Two solve above equations we have to multiply first equation by 2 so we will get;
[tex]\\[/tex]
[tex]\tt2a+24d+36=0 -- EQ (1) [/tex]
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Now from subtracting EQ (1) from EQ(2) we will get;
[tex]\\[/tex]
[tex]\tt 31d - 93 = 0[/tex]
[tex]\tt∴ 31d - 93 = 0[/tex]
[tex]\tt∴ 31d = 93[/tex]
[tex]\tt∴ d = 93/31[/tex]
[tex]\tt∴ d = 3[/tex]
[tex]\\[/tex]
Now putting value of d into EQ(2) we will get..
[tex]\\[/tex]
[tex] \tt 2a + 55 * (3) - 57 = 0[/tex]
[tex] \tt∴ 2a + 165 - 57 = 0[/tex]
[tex] \tt∴ 2a + 108 = 0[/tex]
[tex] \tt∴ 2a = -108 [/tex]
[tex] \tt∴ a = -54 [/tex]
[tex]\\[/tex]
Now we have values of a = -54 and d = 30 so we can easily find S100 as below
[tex]\\[/tex]
[tex] \tt{S_{100}= 100/2 * (2(-54)+(100-1)3)}[/tex]
[tex]\tt∴ S_{100} = 50 * (-108+99*3)[/tex]
[tex]\tt∴ S_{100} = 50 * (-108+297)[/tex]
[tex]\tt∴ S_{100} = 50 * (189)[/tex]
[tex]\tt∴ S_{100} = 50 * (189)[/tex]
[tex]\tt∴ S_{100} = 9450[/tex]
[tex]\\[/tex]
So the Final Answer is,
[tex]\large\tt{S_{100} = \orange{9450}}[/tex]
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#Hope It helps :)
✍ BADGX
