Advertisements
Advertisements
प्रश्न
If the first term of an AP is –5 and the common difference is 2, then the sum of the first 6 terms is ______.
विकल्प
0
5
6
15
Advertisements
उत्तर
If the first term of an AP is –5 and the common difference is 2, then the sum of the first 6 terms is 0.
Explanation:
Given,
a = –5
And d = 2
∴ S6 = `6/2[2a + (6 - 1)d]` ...`[∵ S_n = n/2[2a + (n - 1)d]]`
= 3[2(–5) + 5(2)]
= 3(–10 + 10)
= 0
APPEARS IN
संबंधित प्रश्न
If the term of m terms of an A.P. is the same as the sum of its n terms, show that the sum of its (m + n) terms is zero
Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.
Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.
Which term of the AP 21, 18, 15, …… is -81?
Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.
Find the three numbers in AP whose sum is 15 and product is 80.
Write an A.P. whose first term is a and the common difference is d in the following.
a = 10, d = 5
Find the first term and common difference for the A.P.
0.6, 0.9, 1.2,1.5,...
Find the first term and common difference for the A.P.
127, 135, 143, 151,...
Rs 1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find interest amount after 20 years. For this complete the following activity.
The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P.
Which term of the sequence 114, 109, 104, ... is the first negative term?
If Sn denote the sum of n terms of an A.P. with first term a and common difference dsuch that \[\frac{Sx}{Skx}\] is independent of x, then
Q.4
If the sum of the first four terms of an AP is 40 and that of the first 14 terms is 280. Find the sum of its first n terms.
If an = 3 – 4n, show that a1, a2, a3,... form an AP. Also find S20.
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 five multiples of 3 is ______.
Find the sum:
1 + (–2) + (–5) + (–8) + ... + (–236)
Assertion (A): a, b, c are in A.P. if and only if 2b = a + c.
Reason (R): The sum of first n odd natural numbers is n2.
