Advertisements
Advertisements
प्रश्न
Find the sum of all 11 terms of an A.P. whose 6th term is 30.
Advertisements
उत्तर
Given, 6th term of A.P. = 30
or, a6 = 30
or, a + (6 – 1)d = 30
or, a + 5d = 30 ...(i)
Since, Sum of n terms of A.P. is Sn = `n/2[2a + (n - 1)d]`
∴ S11 = `11/2[2a + (11 - 1)d]`
= `11/2[2d + 10d]`
= `(11 xx 2)/2 [a + 5d]`
= 11 × 30 ...[From equation (i)]
= 330
APPEARS IN
संबंधित प्रश्न
In an AP given d = 5, S9 = 75, find a and a9.
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
How many terms of the A.P. 63, 60, 57, ... must be taken so that their sum is 693?
How many numbers are there between 101 and 999, which are divisible by both 2 and 5?
Which term of the AP 21, 18, 15, … is zero?
Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.
In an A.P., the first term is 22, nth term is −11 and the sum to first n terms is 66. Find n and d, the common difference
If the sum of n terms of an A.P. be 3n2 + n and its common difference is 6, then its first term is ______.
If the second term and the fourth term of an A.P. are 12 and 20 respectively, then find the sum of first 25 terms:
In an A.P., if Sn = 3n2 + 5n and ak = 164, find the value of k.
