Question

find the sums given bellow: 34+32+30+...+10​

Answer 1

Step-by-step explanation:

If u observe this is an arithmetic progression

First find number of terms which is 13 from 34 to 10 go on reducing by 2 and count the number of terms

n=13

first term is 34

a=34

Common difference that is difference between two consecutive numbers is 2

d=2

Last term is 10 which is. l=10

Sum of n terms is given by

Sum=(n/2)(a+l)

Sum=(13/2)(34+10)

Sum=(13/2)(44)

Sum=286

Answer 2

Answer is 286

Explanation :

Here,

a=34

d= -2

34+32+30+...+10

It is in the form of AP...

So,

SUM of numbers in AP

= $\frac{n}{2} {[2a+(n-1)d] }$ ____[i]

where a is first term, d is common difference and n is the term upto which we have to find the sum...

But, n is unknown...

The last term is 10 which is nth term

By formula,

Tn= a+(n-1)d

10=34+(n-1)-2

-24= -2n+2

2n=26

n=13

From [i],

= \sf $\frac{n}{2} {[2a+(n-1)d] } \\\sf </p><p>= \frac{13}{2} {(68-24)} \\\sf </p><p>= \frac{13}{2} x{44} \\\sf </p><p>= \frac{13}{ \cancel{2}}x { \cancel{44}\:\:22}\\\sf </p><p>=13 x 22</p><p>= 286$