Question

Find the sum of A.P. given below: 34+32+30+.........+10​

Answer 1

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286

step-by-step explanation:


♣ What is an A.P

✍️In mathematics, an arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

✍️ Difference here means the second minus the first. For instance, the sequence

5, 7, 9, 11, 13, 15.........


is an arithmetic progression with common difference of 2.



Now,

Given,

the A.P series is

34, 32, 30, 28,...............,10

here,

first term, a = 34

common difference, d = 32-34 = -2

WE KNOW THAT,

nth term of an A.P

is given by,

$a_n$ = a + (n-1)d

=> 10 = 34 - 2 (n-1)

=> 2(n-1) = 24

=> n-1 = 12

=> n = 13

.°. 10 is the 13th term of the given A.P

thus,

there are 13 terms in this A.P

Now,

tk find the sum of ,

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

we k ow that,

sum of n termz of an A.P is given by,

$S_n$ = n/2[2a +(n-1)d]

putting the values of a, n and d,

we get,

$S_n$ = 13/2 [ 68 -24]

= 13× 44/2

= 13 × 22

= 286

HENCE,

SUM OF THE TERMS OF GIVEN A.P IS 286

Answer 2

Answer:

a= 34 , d = -2 , an = 10

10 = 34+(n-1)-2

10 =34 - 2n + 2

10 =34 +2- 2n

10 = 36 -2n

10 - 36 = -2n

-26 = -2n

n = 26/3

n= 13

Sn = n/2. ( 2a + ( n - 1 ) d )

Sn13 = 13 / 2 ( 2 (34) + (13-1)-2)

= 3/2 (68+ (-24)

=13/2 (44)

=13 x 22

= 286