Advertisements
Advertisements
Question
Find the sum of the following arithmetic progressions:
−26, −24, −22, …. to 36 terms
Solution
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
RELATED QUESTIONS
Find the sum of n terms of an A.P. whose nth terms is given by an = 5 − 6n.
Find the sum of the following Aps:
i) 2, 7, 12, 17, ……. to 19 terms .
Choose the correct alternative answer for the following question .
What is the sum of the first 30 natural numbers ?
The A.P. in which 4th term is –15 and 9th term is –30. Find the sum of the first 10 numbers.
Find the sum (−5) + (−8)+ (−11) + ... + (−230) .
The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P.
If the sum of first n terms of an A.P. is \[\frac{1}{2}\] (3n2 + 7n), then find its nth term. Hence write its 20th term.
In an A.P. a = 2 and d = 3, then find S12
Yasmeen saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month. If she continues to save in this manner, in how many months will she save Rs 2000?
Solve the equation
– 4 + (–1) + 2 + ... + x = 437
