Advertisements
Advertisements
Question
First term and the common differences of an A.P. are 6 and 3 respectively; find S27.
Solution: First term = a = 6, common difference = d = 3, S27 = ?
Sn = `"n"/2 [square + ("n" - 1)"d"]` - Formula
Sn = `27/2 [12 + (27 - 1)square]`
= `27/2 xx square`
= 27 × 45
S27 = `square`
Advertisements
Solution
It is given that,
First term (a) = 6
Common difference (d) = 3
Sn = `n/2 [bb(2a) + (n - 1)d]` - Formula
∴ S27 = `27/2 [2(6) + (27 - 1)bb((3))]`
= `27/2 (12 + 26 (3))`
= `27/2 (12 + 78)`
= `27/2 xx bb90`
= 1215
Hence, S27 = 1215.
RELATED QUESTIONS
Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 ×(5 + 3)]
If (m + 1)th term of an A.P is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n + 1)th term.
Find the sum of the following arithmetic progressions:
1, 3, 5, 7, ... to 12 terms
Find the sum of the following arithmetic progressions:
41, 36, 31, ... to 12 terms
Find the sum 25 + 28 + 31 + ….. + 100
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 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.
A sum of ₹2800 is to be used to award four prizes. If each prize after the first is ₹200 less than the preceding prize, find the value of each of the prizes
The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P.
Find the sum: 1 + 3 + 5 + 7 + ... + 199 .
A piece of equipment cost a certain factory Rs 60,000. If it depreciates in value, 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?
The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by \[\frac{l^2 - a^2}{k - (l + a)}\] , then k =
Q.17
The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.
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 sum of first ten natural number is ______.
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.
Kanika was given her pocket money on Jan 1st, 2008. She puts Rs 1 on Day 1, Rs 2 on Day 2, Rs 3 on Day 3, and continued doing so till the end of the month, from this money into her piggy bank. She also spent Rs 204 of her pocket money, and found that at the end of the month she still had Rs 100 with her. How much was her pocket money for the month?
Solve the equation
– 4 + (–1) + 2 + ... + x = 437
