Advertisements
Advertisements
प्रश्न
Find the sum of the following arithmetic progressions:
−26, −24, −22, …. to 36 terms
उत्तर
In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,
`S_n = n/2 [2a + (n -1)d]`
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
−26, −24, −22, …. to 36 terms
Common difference of the A.P. (d) = `a_2 - a_1`
= (-24) - (-26)
= - 24 + 26
= 2
Number of terms (n) = 36
The first term for the given A.P. (a) = −26
So, using the formula we get,
`S_36 = 36/2 [2(-26) + (36 - 1)(2)]`
= (18)[-52 + (35) (2)]
= (18)[-52 + 70]
= (18)[18]
= 324
Therefore the sum of first 36 terms for the given A.P is 324
APPEARS IN
संबंधित प्रश्न
If Sn1 denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 − S4).
If the pth term of an A. P. is `1/q` and qth term is `1/p`, prove that the sum of first pq terms of the A. P. is `((pq+1)/2)`.
Find the sum of the first 51 terms of the A.P: whose second term is 2 and the fourth term is 8.
Find the sum of the first 25 terms of an A.P. whose nth term is given by an = 2 − 3n.
If the sum of first m terms of an AP is ( 2m2 + 3m) then what is its second term?
If the ratio of sum of the first m and n terms of an AP is m2 : n2, show that the ratio of its mth and nth terms is (2m − 1) : (2n − 1) ?
Find the sum: 1 + 3 + 5 + 7 + ... + 199 .
Let the four terms of the AP be a − 3d, a − d, a + d and a + 3d. find A.P.
Find S10 if a = 6 and d = 3
An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.
