Question

the sum of the first three consecutive terms of an AP is 9 and sum of their squares is 35 then sum to n terms of the series is

Answer 1

HELLO DEAR,

let the three consecutive no. be (a-d),a,(a+d)



given that:-

a-d+a+a-d=9

3a=9

a=9/3

a=3------------------(1)


and also


given that:-

(a-d)²+a²+(a+d)²=35


(3-d)²+3²+(3+d)²=35 -----using (1)



9 + d² - 6d + 9 + 9 + d² + 6d=35


27+ 2d²=35

2d² =35-27

2d² = 8

d² =4

d²=+- 2



when d=2


we get,

(3-2),3,(3+2)


=>1,3,5

then the
$t_{n} = (a + (n - 1)d) \\ = > 1 + 2n - 2 \\ = > (2n - 1)$


when d= -2



(3+2),3,(3-2)


=>5,3,1

then the
$t_{n} = 5 + (n - 1)( - 2) \\ \\ = > 5 - 2n + 2 \\ = > (- 2n + 7)$
I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

ANSWER
..........



3-d)²+3²+(3+d)²=35



9 + d² - 6d + 9 + 9 + d² + 6d=35


27+ 2d²=35

2d² =35-27

2d² = 8

d² =4

d²=+- 2







d=2


we get,

(3-2),3,(3+2)


=>1,3,5





}t_{n} = (a + (n - 1)d)
_______________

= > 1 + 2n - 2 \\ = > (2n - 1
___________________
}​t​n​​=(a+(n−1)d)​=>1+2n−2​=>(2n−1)​​ 


when d= -2



(3+2),3,(3-2)


=>5,3,1





×××××××××*




_{n} = 5 + (n - 1)( - 2) \\
________________
= > 5 - 2n + 2 \\ = > (- 2n + 7)
_____________________

​=5+(n−1)(−2)​​=>5−2n+2​=>(−2n+7)​​ 




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