Advertisements
Advertisements
प्रश्न
Find the sum:
`4 - 1/n + 4 - 2/n + 4 - 3/n + ...` upto n terms
उत्तर
Here, first term, a = `4 - 1/n`
Common difference,
d = `(4 - 2/n) - (4 - 1/n)`
= `(-2)/n + 1/n`
= `(-1)/n`
∵ Sum of n terms of an AP,
Sn = `n/2[2a + (n - 1)d]`
⇒ Sn = `n/2[2(4 - 1/n) + (n - 1)((-1)/n)]`
= `n/2{8 - 2/n - 1 + 1/n}`
= `n/2(7 - 1/n)`
= `n/2 xx ((7n - 1)/n)`
= `(7n - 1)/2`
APPEARS IN
संबंधित प्रश्न
The houses in a row numbered consecutively from 1 to 49. Show that there exists a value of x such that sum of numbers of houses preceding the house numbered x is equal to sum of the numbers of houses following x.
If Sn1 denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 − S4).
Which term of the progression 20, 19`1/4`,18`1/2`,17`3/4`, ... is the first negative term?
Find the sum of the following Aps:
i) 2, 7, 12, 17, ……. to 19 terms .
In an A.P. a = 2 and d = 3, then find S12
The sum of first five multiples of 3 is ______.
How many terms of the AP: –15, –13, –11,... are needed to make the sum –55? Explain the reason for double answer.
Find the sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
Find the sum of all even numbers from 1 to 250.
If the first term of an A.P. is 5, the last term is 15 and the sum of first n terms is 30, then find the value of n.
