Question

the sum of three consecutive terms of ap is 9 and sum of their product 35

Answer 1

SORRY
the sum of three consecutive terms of ap is 9 and sum of their squares is 35
find \[S _{n}\]

Answer 2

Can you write the three terms of the ap?

Answer 3

a-d,a,a+d

Answer 4

Let the 3 terms be (a-d),(a) and (a+d)

Here, a-d+a+a+d = 3a = 9 , or a = 3 ..
Similarly, do for their squares, and then, do tell what do you get!

Answer 5

d=2

Answer 6

then Sn is ..

Answer 7

@wwe123 Do you understand anything ?

Answer 8

i understand well and also able to solve the question,but Answer not match

Answer 9

Can you show me your working so i can tell you where you went wrong ?

Answer 10

1 min

Answer 11

a-d+a+a+d= 9
3a=9
a=3

Answer 12

please continue ?

Answer 13

You know the Sn formula, no ?
n/2(2a+(n-1)d)

Answer 14

(a-d)^2+a^2+(a-d)^2=35
a^2+d^2-2ad+a^2+a^2+d^2+2ad=35
3a^2+2d^2=35
2d^2=35-27
2d^2=8
d^2=4
d=2

Answer 15

Now you know "a" and "d" , put these in the Sn formula .. Don't you get the answer that way ?? :/

Answer 16

i get S=n(2+n)

Answer 17

@hba

Answer 18

but answer is n^2 or n(6-n)

Answer 19

well, the terms are
1,3,5
for which neither n^2 nor n(6-n) is correct

Answer 20

S= n/2(2*(3-2)+(n-1)*2) = n(1+n-1) = n(n) = n^2

Answer 21

i think given anwer is wrong
becoz
Sn=n^2 when a=2,d=2

Answer 22

the first term is a-d = 3-2 = 1
then u get n^2.

Answer 23

yes

Answer 24

with d^2 = 4, u also get d=-2

Answer 25

S= n/2(2*(3+2)+(n-1)*(-2)) = n(5-n+1) = n(6-n)
there u go....

Answer 26

here d was -2 so , first term = a-d = 3-(-2) = 3+2