Question

If the 35th term of an A.P is 30, prove that the sum of its 69 terms will be 2070

Answer 1

actually, just multiplying 30 by 69 :V

Answer 2

use the n'th term formula
\(\Large a_n = a_1 +(n-1)d\)

plug in n=35
\(a_{35} = 30\)

you will get an equation in a1 and 'd'

Then use the sum formula,
\(\Large S_n = (n/2)(2a_1 +(n-1)d)\)
and use the equation you got before.

Answer 3

yes, it would be same as multiplying 30 by 69.
if you follow that process, you'll know why :)

Answer 4

dont hoped, i was just guessing :)

Answer 5

that happened because n-1 = 34
and 2(n-1) = 68
which is also 69-1

Answer 6

ohhh yes, i checked it that like i thought :P
S69 = 69/2 (2a1 + 68) = 69/2 * 2 * (a1 + 34) = 69 * 30 = 2070

Answer 7

\(\huge \checkmark \)

Answer 8

Rakesh, ask if any doubts? :)

Answer 9

69*30 HOW IT COMES

Answer 10

did you follow the steps in my first comment ?
where are you stuck ?

Answer 11

a1+34 = 30 cant get

Answer 12

you missed the 'd'
\(\Large 30 = a_1 +34 d \quad \to (1)\)

thats your 1st equation

Answer 13

now use the sum formula

Answer 14

S69 = 69/2 (2a1 + 68) = 69/2 * 2 * (a1 + 34) = 69 * 30 = 2070 wanna understand this only

Answer 15

did you try plugging in values in the sum formula by yourself ?

Answer 16

forget that you ever saw that line

Answer 17

ok

Answer 18

could you complete the problem ?
\(\Large S_{69} = ... ?\)

Answer 19

I UNDERSTAND D FORMULA

Answer 20

only d problem with that 30

Answer 21

s69=69/2(2a+68), = 69/2*2(a+34), 69*(a+34)

Answer 22

you again missed the 'd'


s69=69/2(2a+68\(\Large d\)),

= 69/2*2(a+34\(\Large d\)),

=69*(a+34\(\Large d\))

and using equation (1),
= 69*30
=2070