Advertisements
Advertisements
प्रश्न
Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.
Advertisements
उत्तर
The odd integers between 1 and 1000 that are divisible by 3 are:
3, 9, 15, 21...999
Here, we have:
\[a = 3, d = 6\]
\[ a_n = 999\]
\[ \Rightarrow 3 + (n - 1)6 = 999\]
\[ \Rightarrow 3 + 6n - 6 = 999\]
\[ \Rightarrow 6n = 1002\]
\[ \Rightarrow n = 167\]
\[ S_n = \frac{n}{2}\left[ 2a + (n - 1)d \right]\]
\[ \Rightarrow S_{167} = \frac{167}{2}\left[ 2 \times 3 + (167 - 1)6 \right]\]
\[ \Rightarrow S_{167} = \frac{167}{2}\left[ 1002 \right] = 83667\]
\[\text { Hence, proved } .\]
APPEARS IN
संबंधित प्रश्न
How many terms of the A.P. -6 , `-11/2` , -5... are needed to give the sum –25?
Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)th numbers is 5:9. Find the value of m.
A man starts repaying a loan as first installment of Rs. 100. If he increases the installment by Rs 5 every month, what amount he will pay in the 30th installment?
Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3 (S2– S1)
The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.
if `a(1/b + 1/c), b(1/c+1/a), c(1/a+1/b)` are in A.P., prove that a, b, c are in A.P.
Let < an > be a sequence defined by a1 = 3 and, an = 3an − 1 + 2, for all n > 1
Find the first four terms of the sequence.
Let < an > be a sequence. Write the first five term in the following:
a1 = a2 = 2, an = an − 1 − 1, n > 2
The nth term of a sequence is given by an = 2n2 + n + 1. Show that it is not an A.P.
\[\text { If } \theta_1 , \theta_2 , \theta_3 , . . . , \theta_n \text { are in AP, whose common difference is d, then show that }\]
\[\sec \theta_1 \sec \theta_2 + \sec \theta_2 \sec \theta_3 + . . . + \sec \theta_{n - 1} \sec \theta_n = \frac{\tan \theta_n - \tan \theta_1}{\sin d} \left[ NCERT \hspace{0.167em} EXEMPLAR \right]\]
If the sum of three numbers in A.P. is 24 and their product is 440, find the numbers.
Find the sum of the following arithmetic progression :
1, 3, 5, 7, ... to 12 terms
Find the sum of the following arithmetic progression :
41, 36, 31, ... to 12 terms
Find the sum of all integers between 84 and 719, which are multiples of 5.
Find the sum of all even integers between 101 and 999.
Solve:
1 + 4 + 7 + 10 + ... + x = 590.
Find the r th term of an A.P., the sum of whose first n terms is 3n2 + 2n.
Find the sum of n terms of the A.P. whose kth terms is 5k + 1.
Find the sum of odd integers from 1 to 2001.
If S1 be the sum of (2n + 1) terms of an A.P. and S2 be the sum of its odd terms, then prove that: S1 : S2 = (2n + 1) : (n + 1).
If \[\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\] are in A.P., prove that:
\[\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}\] are in A.P.
If \[\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\] are in A.P., prove that:
a (b +c), b (c + a), c (a +b) are in A.P.
If a, b, c is in A.P., then show that:
a2 (b + c), b2 (c + a), c2 (a + b) are also in A.P.
If a, b, c is in A.P., prove that:
a2 + c2 + 4ac = 2 (ab + bc + ca)
A man accepts a position with an initial salary of ₹5200 per month. It is understood that he will receive an automatic increase of ₹320 in the very next month and each month thereafter.
(i) Find his salary for the tenth month.
(ii) What is his total earnings during the first year?
A man saved ₹66000 in 20 years. In each succeeding year after the first year he saved ₹200 more than what he saved in the previous year. How much did he save in the first year?
Write the common difference of an A.P. the sum of whose first n terms is
Write the sum of first n even natural numbers.
Write the value of n for which n th terms of the A.P.s 3, 10, 17, ... and 63, 65, 67, .... are equal.
If \[\frac{3 + 5 + 7 + . . . + \text { upto n terms }}{5 + 8 + 11 + . . . . \text { upto 10 terms }}\] 7, then find the value of n.
If n arithmetic means are inserted between 1 and 31 such that the ratio of the first mean and nth mean is 3 : 29, then the value of n is
Let Sn denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = Sn − k Sn − 1 + Sn − 2 , then k =
Mark the correct alternative in the following question:
The 10th common term between the A.P.s 3, 7, 11, 15, ... and 1, 6, 11, 16, ... is
Mark the correct alternative in the following question:
\[\text { If in an A . P } . S_n = n^2 q \text { and } S_m = m^2 q, \text { where } S_r \text{ denotes the sum of r terms of the A . P . , then }S_q \text { equals }\]
The first term of an A.P.is a, and the sum of the first p terms is zero, show that the sum of its next q terms is `(-a(p + q)q)/(p - 1)`
If the sum of n terms of an A.P. is given by Sn = 3n + 2n2, then the common difference of the A.P. is ______.
The sum of terms equidistant from the beginning and end in an A.P. is equal to ______.
The number of terms in an A.P. is even; the sum of the odd terms in lt is 24 and that the even terms is 30. If the last term exceeds the first term by `10 1/2`, then the number of terms in the A.P. is ______.
If b2, a2, c2 are in A.P., then `1/(a + b), 1/(b + c), 1/(c + a)` will be in ______
