Advertisements
Advertisements
प्रश्न
Find the sum of the following APs.
`1/15, 1/12, 1/10`, ......, to 11 terms.
उत्तर
`1/15, 1/12, 1/10,` ......, to 11 terms
For this A.P.,
a = `1/15`
n = 11
d = a2 - a1
= `1/12-1/15`
= `(5-4)/60`
= `1/60`
We know that
Sn = `n/2[2a + (n -1)d]`
S11 = `11/2[2(1/15)+(11-1)1/60]`
S11 = `11/2[2/5+10/60]`
S11 = `11.2[2/15+1/6]`
S11 = `11/2[(4+5)/30]`
S11 = `(11/2)(9/30)`
S11 = `33/20`
Thus, the required sum of first 11 terms is `33/20`.
APPEARS IN
संबंधित प्रश्न
The ratio of the sum use of n terms of two A.P.’s is (7n + 1) : (4n + 27). Find the ratio of their mth terms
If the sum of first m terms of an A.P. is the same as the sum of its first n terms, show that the sum of its first (m + n) terms is zero
Three numbers are in A.P. If the sum of these numbers is 27 and the product 648, find the numbers.
Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.
Determine k so that (3k -2), (4k – 6) and (k +2) are three consecutive terms of an AP.
If k,(2k - 1) and (2k - 1) are the three successive terms of an AP, find the value of k.
If the sum of first m terms of an AP is ( 2m2 + 3m) then what is its second term?
What is the 5th term form the end of the AP 2, 7, 12, …., 47?
The sum of the first n terms of an AP in `((5n^2)/2 + (3n)/2)`.Find its nth term and the 20th term of this AP.
The A.P. in which 4th term is –15 and 9th term is –30. Find the sum of the first 10 numbers.
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 5th and 9th terms of an A.P. is 30. If its 25th term is three times its 8th term, find the A.P.
The sum of first 9 terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1 : 2. Find the first and 15th term of the A.P.
In an A.P., the sum of first ten terms is −150 and the sum of its next ten terms is −550. Find the A.P.
If Sn denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 − S4).
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 − kSn−1 + Sn−2, then k =
If in an A.P. Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to
The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its
The common difference of the A.P. is \[\frac{1}{2q}, \frac{1 - 2q}{2q}, \frac{1 - 4q}{2q}, . . .\] is
Q.7
Find the sum of three-digit natural numbers, which are divisible by 4
A merchant borrows ₹ 1000 and agrees to repay its interest ₹ 140 with principal in 12 monthly instalments. Each instalment being less than the preceding one by ₹ 10. Find the amount of the first instalment
Find t21, if S41 = 4510 in an A.P.
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.
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.
Find the sum:
`(a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) +` ... to 11 terms
In an A.P., the sum of first n terms is `n/2 (3n + 5)`. Find the 25th term of the A.P.
The nth term of an Arithmetic Progression (A.P.) is given by the relation Tn = 6(7 – n)..
Find:
- its first term and common difference
- sum of its first 25 terms
The nth term of an A.P. is 6n + 4. The sum of its first 2 terms is ______.
