Question

find the sums given below 34+32+30+......+10​

Answer 1

they are in AP with a=34,d=-2 so

a+(n-1)d=10

n=10-34/-2 +1

n=13

sum of all =n/2[2a+(n-1)d]

=13/2[64-24]

=13/2(40)

=13*20

=260

Answer 2

So, We are required to find the sum of given series. Evaluating this 34+32+30+......+10 directly takes lot of time. So, We try to see if there's something to solve this.

The first term is 34, Second term is 32, Third term is 30. So there's a definite common difference.

Concluding that, This is more like an A. P ( Arithmetic Progression).

First term (a) = 34

Second term (a2) = 32

Last term (l) = 10

We know that,

$d = a_{n + 1} - a_{n}$

So,

$d = a_{2}- a = 32 - 34 = - 2$

Using the formula for sum of an A. P with First and Last terms being a, l respectively.

$S = \boxed{ \frac{n}{2} (a + l)}$

We have everything required except the number of terms.

So, We consider this AP has m terms and mth term is 10

That implies,

10 = 34 + ( m - 1) - 2

10 - 34 = (m-1)-2

-24 = (m - 1) × - 2

12 = m - 1

12 + 1 = m

m = 13

Now, Use the sum formula,

34+32+30+......+10

= 13/2 ( 34 + 10).

= 13/2 ( 44)

= 13(22)

= 286

Therefore, 34+32+30+......+10 = 286