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Ask QuestionPosted by Saket S 6 years, 3 months ago
- 1 answers
Yogita Ingle 6 years, 3 months ago
Formula of first n terms in AP = {tex}S_n = \frac{n}{2}(2a+(n-1)d){/tex}
Substitute n = 7
So,{tex} S_7 = \frac{7}{2}(2a+6d){/tex}
{tex}49 = \frac{7}{2}(2a+6d){/tex}
{tex}49 \times \frac{2}{7} = 2a+6d{/tex}
14= 2a+6d
Substitute n = 17
{tex}S_{17} = \frac{17}{2}(2a+16d){/tex}
{tex}289 = \frac{17}{2}(2a+16d){/tex}
{tex}289 \times \frac{2}{17} = 2a+16d{/tex}
34= 2a+16d
34= 2a+10d+6d
Using A
34= 14+10d
20 = 10d
d=2
Substitute the value of d in A
14= 2a+12
2= 2a
a=1
So, {tex} S_n = \frac{n}{2}(2(1)+(n-1)2){/tex}
{tex}S_n = \frac{n}{2}(2+(n-1)2){/tex}
Posted by Hoodie Protocol 6 years, 3 months ago
- 1 answers
Yogita Ingle 6 years, 3 months ago
The given points are (7, 2) and (-1, 0)
The slope of the line is given by
{tex}m=\frac{y_2-y_1}{x_2-x-1}{/tex}
Substituting the values
{tex}m=\frac{0-2}{-1-7}\\\\m=\frac{-2}{-8}\\\\m=\frac{1}{4}{/tex}
Point slope form of a line is given by
{tex}y-y_1=m(x-x_1)\\\\y-2=\frac{1}{4}(x-7)\\\\4y-8=x-7\\\\4y=x+1{/tex}
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- 0 answers
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- 1 answers
Yogita Ingle 6 years, 3 months ago
First term =a
second term =b
last term =c
common difference=(b-a)
tn =a+(n-1) d
c=a+(n-1) d
(c-a)=(n-1)(b-a)
(c-a)/(b-a)+1=n
( c+b-2a)/(b-a)=n--------------------(1)
now ,
we know sum of n terms=n/2 (first term+last term)
put equation (1)value
= (b+c-2a)/(b-a)(a+c)
hence Sn=[(b+c-2a)(c+a)/(b-a)]
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Aadya Singh ? 6 years, 3 months ago
2Thank You