Advertisements
Advertisements
प्रश्न
Find the sum of the following APs.
0.6, 1.7, 2.8, …….., to 100 terms.
Advertisements
उत्तर
0.6, 1.7, 2.8, …, to 100 terms
For this A.P.,
a = 0.6
d = a2 − a1
= 1.7 − 0.6
d = 1.1
n = 100
We know that
Sn = `n/2[2a+(n-1)d]`
S100 = `100/2[2(0.6)+(100 - 1)1.1]`
= 50[1.2 + (99) × (1.1)]
= 50[1.2 + 108.9]
= 50[110.1]
= 5505
Thus, the required sum of first 100 terms is 5505.
संबंधित प्रश्न
Find the 9th term from the end (towards the first term) of the A.P. 5, 9, 13, ...., 185
If Sn1 denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 − S4).
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 all integers between 100 and 550, which are divisible by 9.
In an A.P., if the first term is 22, the common difference is −4 and the sum to n terms is 64, find n.
How many terms of the A.P. : 24, 21, 18, ................ must be taken so that their sum is 78?
Find the sum of all odd natural numbers less than 50.
The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term.
Write an A.P. whose first term is a and common difference is d in the following.
Find the first term and common difference for the following A.P.:
5, 1, –3, –7, ...
In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms.
(Assume that three consecutive terms in A.P. are a – d, a, a + d).
The sum of first seven terms of an A.P. is 182. If its 4th and the 17th terms are in the ratio 1 : 5, find the A.P.
If Sn denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 − S4).
Which term of the sequence 114, 109, 104, ... is the first negative term?
If `4/5` , a, 2 are three consecutive terms of an A.P., then find the value of a.
If Sr denotes the sum of the first r terms of an A.P. Then , S3n: (S2n − Sn) is
Q.19
Determine the sum of first 100 terms of given A.P. 12, 14, 16, 18, 20,......
Activity :- Here, a = 12, d = `square`, n = 100, S100 = ?
Sn = `"n"/2 [square + ("n" - 1)"d"]`
S100 = `square/2 [24 + (100 - 1)"d"]`
= `50(24 + square)`
= `square`
= `square`
Shubhankar invested in a national savings certificate scheme. In the first year he invested ₹ 500, in the second year ₹ 700, in the third year ₹ 900 and so on. Find the total amount that he invested in 12 years.
