Advertisements
Advertisements
Question
Find the sum of the following arithmetic progressions: 50, 46, 42, ... to 10 terms
Advertisements
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
50, 46, 42, ... to 10 terms
Common difference of the A.P. (d)
`= a_2 - a_1`
= 46 - 50
= -4
Number of terms (n) = 10
First term for the given A.P. (a) = 50
So using the formula we get
`S_10 = 10/2 [2(50) + (10 - 1)(-4)]`
= (5)[100 + (9)(-4)]
= (5)[100 - 36]
= (5)[64]
= 320
Therefore the sum of first 10 terms for the given A.P is 320
APPEARS IN
RELATED QUESTIONS
How many terms of the A.P. 65, 60, 55, .... be taken so that their sum is zero?
Check whether -150 is a term of the A.P. 11, 8, 5, 2, ....
In an AP given d = 5, S9 = 75, find a and a9.
Find the 12th term from the end of the following arithmetic progressions:
3, 5, 7, 9, ... 201
The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.
The 4th term of an AP is zero. Prove that its 25th term is triple its 11th term.
Choose the correct alternative answer for the following question .
If for any A.P. d = 5 then t18 – t13 = ....
What is the sum of first 10 terms of the A. P. 15,10,5,........?
Find where 0 (zero) is a term of the A.P. 40, 37, 34, 31, ..... .
The sum of the first n terms of an A.P. is 4n2 + 2n. Find the nth term of this A.P.
If Sn denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 − S4).
The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be
If 18th and 11th term of an A.P. are in the ratio 3 : 2, then its 21st and 5th terms are in the ratio
Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
The first term of an AP is –5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.
In an AP, if Sn = n(4n + 1), find the AP.
Find the sum of all odd numbers between 351 and 373.
Find the sum of first 20 terms of an A.P. whose nth term is given as an = 5 – 2n.
Solve the equation:
– 4 + (–1) + 2 + 5 + ... + x = 437
k + 2, 2k + 7 and 4k + 12 are the first three terms of an A.P. The first term of this A.P. is ______.
